Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations Exercise 4.3

Question 1. Solve: \( x \frac { dy }{dx} = x + y \)
Answer: We are given the differential equation: \( x \frac { dy }{dx} = x + y \)
First, we can rearrange this equation to isolate \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x+y}{x} \) ......... (1)
This is a homogeneous differential equation because all terms (x, y) have the same degree (degree 1). To solve it, we use the substitution method.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{x+vx}{x} \)
\( \implies v + x \frac{dv}{dx} = \frac{x(1+v)}{x} \)
\( \implies v + x \frac{dv}{dx} = 1+v \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = 1+v-v \)
\( \implies x \frac{dv}{dx} = 1 \)
Next, separate the variables \(v\) and \(x\):
\( dv = \frac{1}{x} dx \)
Now, we integrate both sides:
\( \int dv = \int \frac{1}{x} dx \)
\( \implies v = \log |x| + c \)
We know that \( \log |x| + c \) can be written as \( \log |x| + \log |e^c| \), which simplifies to \( \log |C x| \) where \( C = e^c \) is an arbitrary constant.
So, \( v = \log |Cx| \).
To remove the logarithm, we can write this in exponential form: \( e^v = Cx \).
From our substitution, we have \( y = vx \), which means \( v = \frac{y}{x} \).
Substitute \(v\) back into the equation:
\( e^{y/x} = Cx \)
Therefore, the solution to the differential equation is \( x = C e^{y/x} \).
In simple words: We changed the given equation into a form where we could separate the variables. By replacing 'y' with 'vx' and then integrating, we found the final relationship between x and y. This method helps solve equations where all terms have the same power.

🎯 Exam Tip: When solving homogeneous differential equations, always remember the substitution \( y = vx \) and then convert \( \frac{dy}{dx} \) to \( v + x \frac{dv}{dx} \). Pay close attention to separating the variables correctly before integration.

Question 2. Solve: \( (x - y) \frac { dy }{dx} = x + 3y \)
Answer: We are given the differential equation: \( (x - y) \frac { dy }{dx} = x + 3y \)
First, rearrange the equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x+3y}{x-y} \) ......... (1)
This is a homogeneous differential equation because all terms (x, y) have the same degree (degree 1). To solve it, we use the substitution method.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{x+3vx}{x-vx} \)
\( \implies v + x \frac{dv}{dx} = \frac{x(1+3v)}{x(1-v)} \)
\( \implies v + x \frac{dv}{dx} = \frac{1+3v}{1-v} \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = \frac{1+3v}{1-v} - v \)
To combine the terms on the right side, find a common denominator:
\( x \frac{dv}{dx} = \frac{1+3v - v(1-v)}{1-v} \)
\( \implies x \frac{dv}{dx} = \frac{1+3v - v + v^2}{1-v} \)
\( \implies x \frac{dv}{dx} = \frac{v^2+2v+1}{1-v} \)
Notice that the numerator is a perfect square: \( v^2+2v+1 = (v+1)^2 \).
So, \( x \frac{dv}{dx} = \frac{(v+1)^2}{1-v} \)
Next, separate the variables \(v\) and \(x\):
\( \frac{1-v}{(v+1)^2} dv = \frac{1}{x} dx \)
We can split the left side into two terms:
\( \frac{1}{(v+1)^2} - \frac{v}{(v+1)^2} dv = \frac{1}{x} dx \)
To integrate \( \frac{v}{(v+1)^2} \), we can write \( v = (v+1)-1 \):
\( \left( \frac{1}{(v+1)^2} - \frac{(v+1)-1}{(v+1)^2} \right) dv = \frac{1}{x} dx \)
\( \implies \left( \frac{1}{(v+1)^2} - \left( \frac{1}{v+1} - \frac{1}{(v+1)^2} \right) \right) dv = \frac{1}{x} dx \)
\( \implies \left( \frac{1}{(v+1)^2} - \frac{1}{v+1} + \frac{1}{(v+1)^2} \right) dv = \frac{1}{x} dx \)
\( \implies \left( \frac{2}{(v+1)^2} - \frac{1}{v+1} \right) dv = \frac{1}{x} dx \)
Now, integrate both sides:
\( \int \left( 2(v+1)^{-2} - \frac{1}{v+1} \right) dv = \int \frac{1}{x} dx \)
\( \implies 2 \frac{(v+1)^{-1}}{-1} - \log |v+1| = \log |x| + \log |c| \)
\( \implies - \frac{2}{v+1} - \log |v+1| = \log |cx| \)
Move the log terms to one side:
\( - \frac{2}{v+1} = \log |cx| + \log |v+1| \)
\( \implies - \frac{2}{v+1} = \log |cx(v+1)| \)
Now, substitute \( v = \frac{y}{x} \) back into the equation:
\( - \frac{2}{\frac{y}{x}+1} = \log \left| cx \left( \frac{y}{x} + 1 \right) \right| \)
Simplify the denominators:
\( - \frac{2}{\frac{y+x}{x}} = \log \left| cx \left( \frac{y+x}{x} \right) \right| \)
\( \implies - \frac{2x}{y+x} = \log |c(y+x)| \)
We can write this in exponential form:
\( e^{- \frac{2x}{x+y}} = c(x+y) \)
So, \( \frac{1}{c} e^{- \frac{2x}{x+y}} = x+y \). Let \( K = \frac{1}{c} \).
The final solution is \( (x+y) = K e^{- \frac{2x}{x+y}} \). This form makes it easy to see the exponential relationship between the two sides.
In simple words: We used a special trick of replacing 'y' with 'vx' because the powers of x and y were the same in all parts of the equation. After doing this, we separated the 'v' terms from the 'x' terms and then solved by integrating both sides. Finally, we put 'y' back in place of 'vx' to get the answer.

🎯 Exam Tip: Homogeneous equations often lead to complex integration. Remember to simplify the algebraic expressions as much as possible before integrating, and use partial fractions or algebraic manipulation for terms like \( \frac{1-v}{(v+1)^2} \).

Question 3. Solve: \( x \frac { dy }{dx} – y = \sqrt { x^2+y^2} \)
Answer: We are given the differential equation: \( x \frac { dy }{dx} – y = \sqrt { x^2+y^2} \)
First, rearrange the equation to find \( \frac{dy}{dx} \):
\( x \frac { dy }{dx} = y + \sqrt { x^2+y^2} \)
\( \implies \frac{dy}{dx} = \frac{y + \sqrt { x^2+y^2}}{x} \) ......... (1)
This is a homogeneous differential equation because all terms (y, \( \sqrt{x^2+y^2} \)) have a degree of 1. To solve it, we use the substitution method.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt { x^2+(vx)^2}}{x} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx + \sqrt { x^2(1+v^2)}}{x} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx + |x|\sqrt { 1+v^2}}{x} \)
Assuming \(x > 0\), we have \( |x| = x \).
\( \implies v + x \frac{dv}{dx} = \frac{x(v + \sqrt { 1+v^2})}{x} \)
\( \implies v + x \frac{dv}{dx} = v + \sqrt { 1+v^2} \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = \sqrt { 1+v^2} \)
Next, separate the variables \(v\) and \(x\):
\( \frac{dv}{\sqrt { 1+v^2}} = \frac{1}{x} dx \)
Now, integrate both sides. The integral of \( \frac{1}{\sqrt{1+v^2}} \) is a standard result, \( \log |v + \sqrt{1+v^2}| \).
\( \int \frac{dv}{\sqrt { 1+v^2}} = \int \frac{1}{x} dx \)
\( \implies \log |v + \sqrt { 1+v^2}| = \log |x| + \log |c| \)
Using logarithm properties \( \log a + \log b = \log (ab) \):
\( \log |v + \sqrt { 1+v^2}| = \log |cx| \)
Since the logarithms are equal, their arguments must be equal:
\( v + \sqrt { 1+v^2} = cx \)
Now, substitute \( v = \frac{y}{x} \) back into the equation:
\( \frac{y}{x} + \sqrt { 1+\left(\frac{y}{x}\right)^2} = cx \)
\( \implies \frac{y}{x} + \sqrt { 1+\frac{y^2}{x^2}} = cx \)
\( \implies \frac{y}{x} + \sqrt { \frac{x^2+y^2}{x^2}} = cx \)
\( \implies \frac{y}{x} + \frac{\sqrt { x^2+y^2}}{|x|} = cx \)
Assuming \(x > 0\), so \( |x| = x \):
\( \implies \frac{y}{x} + \frac{\sqrt { x^2+y^2}}{x} = cx \)
Multiply the entire equation by \(x\):
\( y + \sqrt { x^2+y^2} = cx^2 \)
This is the general solution for the given differential equation. This type of equation often appears in physics problems involving vector magnitudes.
In simple words: We saw that the equation had terms with the same overall power. So, we replaced 'y' with 'vx' to make it easier. After simplifying and doing an integration, we swapped 'vx' back to 'y' to get the final answer.

🎯 Exam Tip: For equations involving \( \sqrt{x^2+y^2} \), the homogeneous substitution \( y=vx \) is particularly effective. Remember the standard integral \( \int \frac{1}{\sqrt{1+v^2}} dv = \log|v+\sqrt{1+v^2}| \).

Question 4. Solve: \( \frac { dy }{dx} = \frac { 3x-2y }{2x-3y} \)
Answer: We are given the differential equation: \( \frac { dy }{dx} = \frac { 3x-2y }{2x-3y} \) ........ (1)
This is a homogeneous differential equation because all terms (x, y) have the same degree (degree 1). To solve it, we use the substitution method.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{3x-2vx}{2x-3vx} \)
\( \implies v + x \frac{dv}{dx} = \frac{x(3-2v)}{x(2-3v)} \)
\( \implies v + x \frac{dv}{dx} = \frac{3-2v}{2-3v} \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = \frac{3-2v}{2-3v} - v \)
To combine the terms on the right side, find a common denominator:
\( x \frac{dv}{dx} = \frac{3-2v - v(2-3v)}{2-3v} \)
\( \implies x \frac{dv}{dx} = \frac{3-2v - 2v + 3v^2}{2-3v} \)
\( \implies x \frac{dv}{dx} = \frac{3v^2-4v+3}{2-3v} \)
Next, separate the variables \(v\) and \(x\):
\( \frac{2-3v}{3v^2-4v+3} dv = \frac{1}{x} dx \)
We can rewrite the numerator to match the derivative of the denominator. The derivative of \( 3v^2-4v+3 \) is \( 6v-4 \). We have \( 2-3v \). We can factor out \( -\frac{1}{2} \) from \( 2-3v \) to get \( -\frac{1}{2} (3v-2) \). This helps with the integration. Oh, wait, the OCR shows \( (6v-4) \) in the next step, let's see how they got there.
Ah, they multiplied by \( -\frac{2}{2} \) to get \( -\frac{1}{2} (6v-4) \). Let's follow their steps directly.
\( \frac{2-3v}{3v^2-4v+3} dv = \frac{1}{x} dx \)
We can write \( (2-3v) = -\frac{1}{2}(6v-4) \).
So, \( -\frac{1}{2} \frac{6v-4}{3v^2-4v+3} dv = \frac{1}{x} dx \)
Now, integrate both sides:
\( -\frac{1}{2} \int \frac{6v-4}{3v^2-4v+3} dv = \int \frac{1}{x} dx \)
The left side is of the form \( \int \frac{f'(v)}{f(v)} dv = \log |f(v)| \). So, \( \int \frac{6v-4}{3v^2-4v+3} dv = \log |3v^2-4v+3| \).
\( \implies -\frac{1}{2} \log |3v^2-4v+3| = \log |x| + \log |c| \)
\( \implies \log |(3v^2-4v+3)^{-1/2}| = \log |cx| \)
Since the logarithms are equal, their arguments must be equal:
\( (3v^2-4v+3)^{-1/2} = cx \)
\( \implies \frac{1}{\sqrt{3v^2-4v+3}} = cx \)
\( \implies 1 = cx \sqrt{3v^2-4v+3} \)
Now, substitute \( v = \frac{y}{x} \) back into the equation:
\( 1 = cx \sqrt{3\left(\frac{y}{x}\right)^2 - 4\left(\frac{y}{x}\right) + 3} \)
\( \implies 1 = cx \sqrt{\frac{3y^2}{x^2} - \frac{4y}{x} + 3} \)
To combine terms under the square root, find a common denominator:
\( 1 = cx \sqrt{\frac{3y^2 - 4xy + 3x^2}{x^2}} \)
\( \implies 1 = cx \frac{\sqrt{3y^2 - 4xy + 3x^2}}{|x|} \)
Assuming \(x > 0\), so \( |x| = x \):
\( \implies 1 = cx \frac{\sqrt{3y^2 - 4xy + 3x^2}}{x} \)
\( \implies 1 = c \sqrt{3y^2 - 4xy + 3x^2} \)
Square both sides to remove the square root:
\( 1^2 = c^2 (3y^2 - 4xy + 3x^2) \)
\( \implies 1 = c^2 (3y^2 - 4xy + 3x^2) \)
Let \( C_1 = c^2 \). Then the solution is \( 1 = C_1 (3y^2 - 4xy + 3x^2) \). This form clearly shows the relationship between x and y.
In simple words: This equation had the same power for all its 'x' and 'y' parts. So, we replaced 'y' with 'vx', simplified, and then separated the 'v' terms and 'x' terms. After integrating, we put 'y' back in place of 'vx' to get the final answer.

🎯 Exam Tip: When integrating \( \frac{f'(v)}{f(v)} \), ensure the numerator is exactly the derivative of the denominator. If not, use a constant multiplier like \( -\frac{1}{2} \) to adjust it. Remember to substitute back \( v = \frac{y}{x} \) at the very end.

Question 5. Solve: \( (y^2 - 2xy) dx = (x^2 - 2xy)dy \)
Answer: We are given the differential equation: \( (y^2 - 2xy) dx = (x^2 - 2xy)dy \)
First, rearrange the equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy} \) ........ (1)
This is a homogeneous differential equation because all terms (y², xy, x²) have the same degree (degree 2). To solve it, we use the substitution method.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{(vx)^2 - 2x(vx)}{x^2 - 2x(vx)} \)
\( \implies v + x \frac{dv}{dx} = \frac{v^2x^2 - 2vx^2}{x^2 - 2vx^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{x^2(v^2 - 2v)}{x^2(1 - 2v)} \)
\( \implies v + x \frac{dv}{dx} = \frac{v^2 - 2v}{1 - 2v} \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = \frac{v^2 - 2v}{1 - 2v} - v \)
To combine the terms on the right side, find a common denominator:
\( x \frac{dv}{dx} = \frac{v^2 - 2v - v(1 - 2v)}{1 - 2v} \)
\( \implies x \frac{dv}{dx} = \frac{v^2 - 2v - v + 2v^2}{1 - 2v} \)
\( \implies x \frac{dv}{dx} = \frac{3v^2 - 3v}{1 - 2v} \)
\( \implies x \frac{dv}{dx} = \frac{3v(v - 1)}{1 - 2v} \)
Next, separate the variables \(v\) and \(x\):
\( \frac{1 - 2v}{3v(v - 1)} dv = \frac{1}{x} dx \)
Now, we need to integrate the left side using partial fractions. Let's decompose \( \frac{1 - 2v}{3v(v - 1)} \):
\( \frac{1 - 2v}{3v(v - 1)} = \frac{A}{v} + \frac{B}{v - 1} \)
\( \implies 1 - 2v = A(3(v - 1)) + B(3v) \)
\( \implies 1 - 2v = 3A(v - 1) + 3Bv \)
Set \( v = 0 \): \( 1 - 0 = 3A(-1) + 0 \implies 1 = -3A \implies A = -\frac{1}{3} \)
Set \( v = 1 \): \( 1 - 2(1) = 0 + 3B(1) \implies -1 = 3B \implies B = -\frac{1}{3} \)
So, \( \frac{1 - 2v}{3v(v - 1)} = \frac{-\frac{1}{3}}{v} + \frac{-\frac{1}{3}}{v - 1} = -\frac{1}{3} \left( \frac{1}{v} + \frac{1}{v - 1} \right) \)
Now, integrate both sides:
\( -\frac{1}{3} \int \left( \frac{1}{v} + \frac{1}{v - 1} \right) dv = \int \frac{1}{x} dx \)
\( \implies -\frac{1}{3} (\log |v| + \log |v - 1|) = \log |x| + \log |c| \)
Using logarithm properties:
\( -\frac{1}{3} \log |v(v - 1)| = \log |cx| \)
\( \implies \log |v(v - 1)|^{-1/3} = \log |cx| \)
\( \implies |v(v - 1)|^{-1/3} = cx \)
\( \implies \frac{1}{(v(v - 1))^{1/3}} = cx \)
Cube both sides:
\( \frac{1}{v(v - 1)} = (cx)^3 \)
\( \implies \frac{1}{v(v - 1)} = c^3 x^3 \)
Now, substitute \( v = \frac{y}{x} \) back into the equation:
\( \frac{1}{\frac{y}{x}\left(\frac{y}{x} - 1\right)} = c^3 x^3 \)
\( \implies \frac{1}{\frac{y}{x}\left(\frac{y - x}{x}\right)} = c^3 x^3 \)
\( \implies \frac{1}{\frac{y(y - x)}{x^2}} = c^3 x^3 \)
\( \implies \frac{x^2}{y(y - x)} = c^3 x^3 \)
Multiply both sides by \( y(y-x) \):
\( x^2 = c^3 x^3 y(y - x) \)
Divide by \( x^2 \) (assuming \( x \neq 0 \)):
\( 1 = c^3 x y(y - x) \)
We can also write this as \( y(y-x) = \frac{1}{c^3 x} \). Let \( K = \frac{1}{c^3} \).
So, the final solution is \( y(y-x) = \frac{K}{x} \) or \( xy(y-x) = K \). The method of partial fractions is very useful here for integrating the 'v' terms.
In simple words: This equation had all its terms with the same total power. We changed 'y' to 'vx' and then separated the 'v' parts from the 'x' parts. We used a special method called partial fractions to integrate the 'v' side. After integration, we changed 'vx' back to 'y' to get the final answer.

🎯 Exam Tip: Homogeneous equations where separation of variables leads to a complex rational function of \(v\) often require partial fraction decomposition for successful integration. Double-check your algebraic manipulations for \(A\) and \(B\) values.

Question 6. The slope of the tangent to a curve at any point (x, y) on it is given by \( (y^3 – 2yx^2) dx + (2xy^2 – x^3) dy = 0 \) and the curve passes through (1, 2). Find the equation of the curve.
Answer: We are given the differential equation for the slope of the tangent:
\( (y^3 – 2yx^2) dx + (2xy^2 – x^3) dy = 0 \)
First, rearrange the equation to find \( \frac{dy}{dx} \):
\( (2xy^2 – x^3) dy = -(y^3 – 2yx^2) dx \)
\( \implies (2xy^2 – x^3) dy = (2yx^2 – y^3) dx \)
\( \implies \frac{dy}{dx} = \frac{2yx^2 – y^3}{2xy^2 – x^3} \) ........ (1)
This is a homogeneous differential equation because all terms (y³, yx², xy², x³) have the same degree (degree 3). To solve it, we use the substitution method.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{2(vx)x^2 – (vx)^3}{2x(vx)^2 – x^3} \)
\( \implies v + x \frac{dv}{dx} = \frac{2vx^3 – v^3x^3}{2vx^3 – x^3} \)
\( \implies v + x \frac{dv}{dx} = \frac{x^3(2v – v^3)}{x^3(2v^2 – 1)} \)
\( \implies v + x \frac{dv}{dx} = \frac{2v – v^3}{2v^2 – 1} \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = \frac{2v – v^3}{2v^2 – 1} - v \)
To combine the terms on the right side, find a common denominator:
\( x \frac{dv}{dx} = \frac{2v – v^3 - v(2v^2 – 1)}{2v^2 – 1} \)
\( \implies x \frac{dv}{dx} = \frac{2v – v^3 - 2v^3 + v}{2v^2 – 1} \)
\( \implies x \frac{dv}{dx} = \frac{3v – 3v^3}{2v^2 – 1} \)
\( \implies x \frac{dv}{dx} = \frac{3v(1 – v^2)}{2v^2 – 1} \)
\( \implies x \frac{dv}{dx} = -\frac{3v(v^2 – 1)}{2v^2 – 1} \)
Next, separate the variables \(v\) and \(x\):
\( \frac{2v^2 – 1}{3v(v^2 – 1)} dv = -\frac{1}{x} dx \)
We can factor \( v^2-1 \) as \( (v-1)(v+1) \). So, \( \frac{2v^2 – 1}{3v(v-1)(v+1)} dv = -\frac{1}{x} dx \)
Now, we need to integrate the left side using partial fractions. Let's decompose \( \frac{2v^2 – 1}{3v(v-1)(v+1)} \):
\( \frac{2v^2 – 1}{3v(v-1)(v+1)} = \frac{A}{v} + \frac{B}{v-1} + \frac{C}{v+1} \)
Multiply by \( 3v(v-1)(v+1) \):
\( 2v^2 – 1 = 3A(v-1)(v+1) + 3Bv(v+1) + 3Cv(v-1) \)
\( \implies 2v^2 – 1 = 3A(v^2-1) + 3B(v^2+v) + 3C(v^2-v) \)
Set \( v = 0 \): \( 2(0)^2 – 1 = 3A(0-1) + 0 + 0 \implies -1 = -3A \implies A = \frac{1}{3} \)
Set \( v = 1 \): \( 2(1)^2 – 1 = 0 + 3B(1)(2) + 0 \implies 1 = 6B \implies B = \frac{1}{6} \)
Set \( v = -1 \): \( 2(-1)^2 – 1 = 0 + 0 + 3C(-1)(-2) \implies 1 = 6C \implies C = \frac{1}{6} \)
So, \( \frac{2v^2 – 1}{3v(v^2 – 1)} = \frac{\frac{1}{3}}{v} + \frac{\frac{1}{6}}{v-1} + \frac{\frac{1}{6}}{v+1} = \frac{1}{3v} + \frac{1}{6(v-1)} + \frac{1}{6(v+1)} \)
Now, integrate both sides:
\( \int \left( \frac{1}{3v} + \frac{1}{6(v-1)} + \frac{1}{6(v+1)} \right) dv = \int -\frac{1}{x} dx \)
\( \implies \frac{1}{3} \log |v| + \frac{1}{6} \log |v-1| + \frac{1}{6} \log |v+1| = -\log |x| + \log |K| \)
\( \implies \log |v^{1/3}| + \log |(v-1)^{1/6}| + \log |(v+1)^{1/6}| = \log |\frac{K}{x}| \)
\( \implies \log |v^{1/3} (v-1)^{1/6} (v+1)^{1/6}| = \log |\frac{K}{x}| \)
\( \implies \log |v^{1/3} ((v-1)(v+1))^{1/6}| = \log |\frac{K}{x}| \)
\( \implies \log |v^{1/3} (v^2-1)^{1/6}| = \log |\frac{K}{x}| \)
\( \implies v^{1/3} (v^2-1)^{1/6} = \frac{K}{x} \)
To remove the fractional exponents, raise both sides to the power of 6:
\( (v^{1/3})^6 ((v^2-1)^{1/6})^6 = \left(\frac{K}{x}\right)^6 \)
\( \implies v^2 (v^2-1) = \frac{K^6}{x^6} \)
Let \( C = K^6 \).
\( v^2 (v^2-1) = \frac{C}{x^6} \)
Now, substitute \( v = \frac{y}{x} \) back into the equation:
\( \left(\frac{y}{x}\right)^2 \left( \left(\frac{y}{x}\right)^2 - 1 \right) = \frac{C}{x^6} \)
\( \implies \frac{y^2}{x^2} \left( \frac{y^2}{x^2} - 1 \right) = \frac{C}{x^6} \)
\( \implies \frac{y^2}{x^2} \left( \frac{y^2 - x^2}{x^2} \right) = \frac{C}{x^6} \)
\( \implies \frac{y^2(y^2 - x^2)}{x^4} = \frac{C}{x^6} \)
Multiply both sides by \( x^6 \):
\( x^2 y^2 (y^2 - x^2) = C \)
This is the general equation of the curve. Now we use the given point (1, 2) to find the value of C.
Substitute \( x = 1 \) and \( y = 2 \) into the equation:
\( (1)^2 (2)^2 ((2)^2 - (1)^2) = C \)
\( \implies 1 \cdot 4 (4 - 1) = C \)
\( \implies 4 \cdot 3 = C \)
\( \implies C = 12 \)
Therefore, the equation of the curve is \( x^2 y^2 (y^2 - x^2) = 12 \). This solution combines differential equations with initial value problems, a common application in calculus.
In simple words: First, we recognized this as a homogeneous equation and used the substitution \( y = vx \). Then, we separated the variables and used a technique called partial fractions to integrate. After solving, we put 'y' back in terms of 'x'. Finally, using the given point (1, 2), we found the specific constant for this curve.

🎯 Exam Tip: When given an initial condition (a point the curve passes through), always remember to substitute those values into the general solution to find the particular constant. Partial fraction decomposition for three terms can be lengthy; check your coefficients carefully.

Question 7. An electric manufacturing company makes small household switches. The company estimates the marginal revenue function for these switches to be \( (x^2 + y^2) dy = xy dx \) where x represents the number of units (in thousands). What is the total revenue function?
Answer: We are given the marginal revenue function as a differential equation:
\( (x^2 + y^2) dy = xy dx \)
First, rearrange the equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \) ......... (1)
This is a homogeneous differential equation because all terms (xy, x², y²) have the same degree (degree 2). To find the total revenue function, we need to solve this differential equation.
Now, let \( y = vx \).
This means that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx^2}{x^2 + v^2x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{x^2 v}{x^2 (1 + v^2)} \)
\( \implies v + x \frac{dv}{dx} = \frac{v}{1 + v^2} \)
Now, subtract \(v\) from both sides:
\( x \frac{dv}{dx} = \frac{v}{1 + v^2} - v \)
To combine the terms on the right side, find a common denominator:
\( x \frac{dv}{dx} = \frac{v - v(1 + v^2)}{1 + v^2} \)
\( \implies x \frac{dv}{dx} = \frac{v - v - v^3}{1 + v^2} \)
\( \implies x \frac{dv}{dx} = \frac{-v^3}{1 + v^2} \)
Next, separate the variables \(v\) and \(x\):
\( \frac{1 + v^2}{-v^3} dv = \frac{1}{x} dx \)
\( \implies -\frac{1 + v^2}{v^3} dv = \frac{1}{x} dx \)
Split the left side into two terms:
\( -\left( \frac{1}{v^3} + \frac{v^2}{v^3} \right) dv = \frac{1}{x} dx \)
\( \implies -\left( v^{-3} + v^{-1} \right) dv = \frac{1}{x} dx \)
Now, integrate both sides:
\( - \int (v^{-3} + v^{-1}) dv = \int \frac{1}{x} dx \)
\( \implies - \left( \frac{v^{-2}}{-2} + \log |v| \right) = \log |x| + \log |c| \)
\( \implies \frac{1}{2v^2} - \log |v| = \log |x| + \log |c| \)
\( \implies \frac{1}{2v^2} = \log |x| + \log |v| + \log |c| \)
\( \implies \frac{1}{2v^2} = \log |cxv| \)
Now, substitute \( v = \frac{y}{x} \) back into the equation:
\( \frac{1}{2\left(\frac{y}{x}\right)^2} = \log \left| cx \left(\frac{y}{x}\right) \right| \)
\( \implies \frac{1}{2\frac{y^2}{x^2}} = \log |cy| \)
\( \implies \frac{x^2}{2y^2} = \log |cy| \)
To express \(y\) explicitly, we can convert the logarithm to an exponential form:
\( e^{\frac{x^2}{2y^2}} = cy \)
\( \implies y = \frac{1}{c} e^{\frac{x^2}{2y^2}} \)
Let \( K = \frac{1}{c} \).
So, the total revenue function is \( y = K e^{\frac{x^2}{2y^2}} \). This shows how the total revenue relates to the number of units produced. Marginal revenue helps companies understand the impact of producing one more unit.
In simple words: We were given an equation for how revenue changes as more items are made. Since the equation had x and y terms with the same powers, we used a trick to simplify it. After separating 'y' and 'x' parts and doing some math called integration, we got the final formula that tells us the total revenue.

🎯 Exam Tip: Remember that marginal revenue is essentially the derivative of total revenue. Solving a marginal revenue differential equation gives the total revenue function. Pay close attention to integrating \( \frac{1}{v} \) as \( \log |v| \) and \( v^{-3} \) as \( \frac{v^{-2}}{-2} \).

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.3

Question 1. Solve: \(x\frac { dy }{dx} = x + y\)
Answer:
Given the differential equation: \(x\frac { dy }{dx} = x + y\)
To solve this, we first rearrange it to isolate \( \frac{dy}{dx} \):
\( \frac { dy }{dx} = \frac { x+y }{x} \) ........ (1)
This equation is a homogeneous differential equation because all terms have the same degree (power of 1 for x and y).
We use the substitution method. Let \( y = vx \).
Then, the derivative of y with respect to x is \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { x+vx }{x} \)
\( v + x\frac { dv }{dx} = \frac { x(1+v) }{x} \)
\( v + x\frac { dv }{dx} = 1+v \)
Now, we separate the variables:
\( x\frac { dv }{dx} = 1+v-v \)
\( x\frac { dv }{dx} = 1 \)
\( dv = \frac { 1 }{x} dx \)
Next, integrate both sides of the equation:
\( \int dv = \int \frac { 1 }{x} dx \)
\( v = \log|x| + c \)
Here, c is the integration constant. We can express x in terms of v:
\( x = e^{v+c} \)
\( x = e^v \cdot e^c \)
Let \( e^c = C \) (another constant).
\( x = C e^v \)
Now, substitute back \( v = \frac{y}{x} \):
\( x = C e^{\frac{y}{x}} \)
This is the general solution for the given differential equation.
In simple words: We changed the given equation into a form where we could separate 'v' and 'x' terms. After doing the integration, we put 'y/x' back in place of 'v' to get the final answer. This type of equation is called homogeneous because all parts have the same power.

🎯 Exam Tip: Remember to clearly state the substitution \(y=vx\) and its derivative \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) at the beginning of solving homogeneous differential equations, as this is a crucial step.

Question 2. Solve: \((x - y) \frac { dy }{dx} = x + 3y\)
Answer:
Given the differential equation: \((x - y) \frac { dy }{dx} = x + 3y\)
First, we need to express \( \frac{dy}{dx} \) clearly:
\( \frac { dy }{dx} = \frac { x+3y }{x-y} \) ........ (1)
This is a homogeneous differential equation because all terms (x, y) are of the same degree (power of 1).
We use the substitution method. Let \( y = vx \).
Then, \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { x+3(vx) }{x-(vx)} \)
\( v + x\frac { dv }{dx} = \frac { x(1+3v) }{x(1-v)} \)
\( v + x\frac { dv }{dx} = \frac { 1+3v }{1-v} \)
Now, separate the variables:
\( x\frac { dv }{dx} = \frac { 1+3v }{1-v} - v \)
\( x\frac { dv }{dx} = \frac { (1+3v) - v(1-v) }{1-v} \)
\( x\frac { dv }{dx} = \frac { 1+3v-v+v^2 }{1-v} \)
\( x\frac { dv }{dx} = \frac { v^2+2v+1 }{1-v} \)
\( x\frac { dv }{dx} = \frac { (v+1)^2 }{1-v} \)
Rearrange to integrate:
\( \frac { 1-v }{(v+1)^2} dv = \frac { 1 }{x} dx \)
To integrate the left side, we can split the fraction:
\( \frac { 1-v }{(v+1)^2} = \frac { (v+1)-2v }{(v+1)^2} = \frac { (v+1)-2(v+1)+2 }{(v+1)^2} = \frac { 1 }{v+1} - \frac { 2 }{ (v+1)^2} \)
So, the equation becomes:
\( \left( \frac { 1 }{v+1} - \frac { 2 }{ (v+1)^2} \right) dv = \frac { 1 }{x} dx \)
Integrate both sides:
\( \int \left( \frac { 1 }{v+1} - 2(v+1)^{-2} \right) dv = \int \frac { 1 }{x} dx \)
\( \log|v+1| - 2\frac { (v+1)^{-1} }{-1} = \log|x| + c \)
\( \log|v+1| + \frac { 2 }{v+1} = \log|x| + c \)
Now substitute back \( v = \frac{y}{x} \):
\( \log\left|\frac{y}{x}+1\right| + \frac { 2 }{\frac{y}{x}+1} = \log|x| + c \)
\( \log\left|\frac{y+x}{x}\right| + \frac { 2x }{y+x} = \log|x| + c \)
\( \log|y+x| - \log|x| + \frac { 2x }{y+x} = \log|x| + c \)
\( \log|y+x| + \frac { 2x }{y+x} = 2\log|x| + c \)
This can also be written as \( \log|y+x| - 2\log|x| + \frac { 2x }{y+x} = c \).
This is the general solution.
In simple words: We solved this equation by replacing 'y' with 'vx' and then separating the variables. After integration, we put 'y/x' back in to find the general answer. The goal is to find a function that satisfies the given rate of change.

🎯 Exam Tip: When integrating fractions involving \( (v+1)^2 \), remember to use partial fractions or a direct power rule for \( (v+1)^{-2} \). Don't forget the constant of integration, 'c', as it's essential for the general solution.

Question 3. Solve: \(x\frac { dy }{dx} – y = \sqrt { x^2+y^2}\)
Answer:
Given the differential equation: \(x\frac { dy }{dx} – y = \sqrt { x^2+y^2}\)
First, we need to isolate \( \frac{dy}{dx} \):
\( x\frac { dy }{dx} = y + \sqrt { x^2+y^2} \)
\( \frac { dy }{dx} = \frac { y + \sqrt { x^2+y^2} }{x} \) ........ (1)
This is a homogeneous differential equation because all terms have a degree of 1 (the square root of \(x^2+y^2\) also gives a degree of 1).
We use the substitution method. Let \( y = vx \).
Then, \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { vx + \sqrt { x^2+(vx)^2} }{x} \)
\( v + x\frac { dv }{dx} = \frac { vx + \sqrt { x^2(1+v^2)} }{x} \)
\( v + x\frac { dv }{dx} = \frac { vx + x\sqrt { 1+v^2} }{x} \)
\( v + x\frac { dv }{dx} = v + \sqrt { 1+v^2} \)
Now, separate the variables:
\( x\frac { dv }{dx} = \sqrt { 1+v^2} \)
\( \frac { dv }{\sqrt { 1+v^2} } = \frac { 1 }{x} dx \)
Integrate both sides. The integral of \( \frac { 1 }{\sqrt { 1+v^2} } \) is \( \log|v + \sqrt{1+v^2}| \).
\( \int \frac { dv }{\sqrt { 1+v^2} } = \int \frac { 1 }{x} dx \)
\( \log|v + \sqrt{1+v^2}| = \log|x| + \log|c| \)
Here, c is the integration constant. We use \( \log|c| \) to simplify the combined logarithm.
\( \log|v + \sqrt{1+v^2}| = \log|cx| \)
Remove the logarithm from both sides:
\( v + \sqrt{1+v^2} = cx \)
Now, substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2} = cx \)
\( \frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = cx \)
\( \frac{y}{x} + \sqrt{\frac{x^2+y^2}{x^2}} = cx \)
\( \frac{y}{x} + \frac{\sqrt{x^2+y^2}}{x} = cx \)
Combine the terms on the left side:
\( \frac{y + \sqrt{x^2+y^2}}{x} = cx \)
Multiply both sides by x:
\( y + \sqrt{x^2+y^2} = cx^2 \)
This is the general solution for the given differential equation. This form represents the family of curves that satisfy the differential equation.
In simple words: We used a special trick called substitution to make the equation simpler, then we separated the x and v parts. After finding the integral of each part, we changed 'v' back to 'y/x' to get the final answer.

🎯 Exam Tip: Recognize the standard integral \( \int \frac{dx}{\sqrt{x^2+a^2}} = \log|x+\sqrt{x^2+a^2}| \) as it appears frequently in these types of problems. Applying it correctly is key to a smooth solution.

Question 4. Solve: \( \frac { dy }{dx} = \frac { 3x-2y }{2x-3y} \)
Answer:
Given the differential equation: \( \frac { dy }{dx} = \frac { 3x-2y }{2x-3y} \) ........ (1)
This is a homogeneous differential equation since all terms (x, y) are of the same degree (power of 1).
We use the substitution method. Let \( y = vx \).
Then, \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { 3x-2(vx) }{2x-3(vx)} \)
\( v + x\frac { dv }{dx} = \frac { x(3-2v) }{x(2-3v)} \)
\( v + x\frac { dv }{dx} = \frac { 3-2v }{2-3v} \)
Now, separate the variables:
\( x\frac { dv }{dx} = \frac { 3-2v }{2-3v} - v \)
\( x\frac { dv }{dx} = \frac { (3-2v) - v(2-3v) }{2-3v} \)
\( x\frac { dv }{dx} = \frac { 3-2v-2v+3v^2 }{2-3v} \)
\( x\frac { dv }{dx} = \frac { 3v^2-4v+3 }{2-3v} \)
Rearrange to integrate:
\( \frac { 2-3v }{3v^2-4v+3} dv = \frac { 1 }{x} dx \)
To integrate the left side, we can notice that the derivative of the denominator \( (3v^2-4v+3) \) is \( (6v-4) \). The numerator is \( (2-3v) \), which is \( -\frac{1}{2}(6v-4) \).
So, multiply the numerator and denominator by -2 on the left side to get the derivative in the numerator:
\( -\frac{1}{2} \int \frac { -2(2-3v) }{3v^2-4v+3} dv = \int \frac { 1 }{x} dx \)
\( -\frac{1}{2} \int \frac { (6v-4) }{3v^2-4v+3} dv = \int \frac { 1 }{x} dx \)
Integrate both sides. The integral of \( \frac{f'(v)}{f(v)} \) is \( \log|f(v)| \).
\( -\frac{1}{2} \log|3v^2-4v+3| = \log|x| + c \)
Multiply by -2:
\( \log|3v^2-4v+3| = -2\log|x| - 2c \)
\( \log|3v^2-4v+3| = \log|x^{-2}| + \log|K| \) (where \( -2c = \log|K| \))
\( \log|3v^2-4v+3| = \log\left|\frac{K}{x^2}\right| \)
Remove logarithms:
\( 3v^2-4v+3 = \frac{K}{x^2} \)
Now, substitute back \( v = \frac{y}{x} \):
\( 3\left(\frac{y}{x}\right)^2 - 4\left(\frac{y}{x}\right) + 3 = \frac{K}{x^2} \)
\( \frac{3y^2}{x^2} - \frac{4y}{x} + 3 = \frac{K}{x^2} \)
Multiply the entire equation by \( x^2 \) to clear the denominators:
\( 3y^2 - 4xy + 3x^2 = K \)
This is the general solution for the given differential equation. The constants simplify beautifully in this case.
In simple words: We used a common trick to solve these equations by replacing 'y' with 'vx'. After rearranging, we found that the top part of the fraction was almost the derivative of the bottom part, which helped us integrate easily using logarithms. Finally, we put 'y/x' back for 'v' to get our answer.

🎯 Exam Tip: When integrating \( \frac{f'(x)}{f(x)} \), remember it simplifies to \( \log|f(x)| \). Look for this pattern in rational functions to simplify the integration process.

Question 5. Solve: \((y² – 2xy) dx = (x² – 2xy)dy\)
Answer:
Given the differential equation: \((y² – 2xy) dx = (x² – 2xy)dy\)
First, we need to express \( \frac{dy}{dx} \):
\( \frac { dy }{dx} = \frac { y^2 - 2xy }{x^2 - 2xy} \) ........ (1)
This is a homogeneous differential equation because all terms have a degree of 2.
We use the substitution method. Let \( y = vx \).
Then, \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { (vx)^2 - 2x(vx) }{x^2 - 2x(vx)} \)
\( v + x\frac { dv }{dx} = \frac { v^2x^2 - 2vx^2 }{x^2 - 2vx^2} \)
\( v + x\frac { dv }{dx} = \frac { x^2(v^2 - 2v) }{x^2(1 - 2v)} \)
\( v + x\frac { dv }{dx} = \frac { v^2 - 2v }{1 - 2v} \)
Now, separate the variables:
\( x\frac { dv }{dx} = \frac { v^2 - 2v }{1 - 2v} - v \)
\( x\frac { dv }{dx} = \frac { (v^2 - 2v) - v(1 - 2v) }{1 - 2v} \)
\( x\frac { dv }{dx} = \frac { v^2 - 2v - v + 2v^2 }{1 - 2v} \)
\( x\frac { dv }{dx} = \frac { 3v^2 - 3v }{1 - 2v} \)
Rearrange to integrate:
\( \frac { 1 - 2v }{3v^2 - 3v} dv = \frac { 1 }{x} dx \)
\( \frac { 1 - 2v }{3v(v-1)} dv = \frac { 1 }{x} dx \)
To integrate the left side, we can use partial fractions. The denominator is \( 3v(v-1) \).
Let \( \frac { 1 - 2v }{3v(v-1)} = \frac { A }{v} + \frac { B }{v-1} \)
Multiply by \( 3v(v-1) \): \( 1 - 2v = 3A(v-1) + 3Bv \)
If \( v=0 \), \( 1 = 3A(-1) \implies A = -\frac{1}{3} \).
If \( v=1 \), \( 1 - 2 = 3B(1) \implies -1 = 3B \implies B = -\frac{1}{3} \).
So, \( \frac { 1 - 2v }{3v(v-1)} = \frac { -\frac{1}{3} }{v} + \frac { -\frac{1}{3} }{v-1} = -\frac{1}{3} \left( \frac{1}{v} + \frac{1}{v-1} \right) \)
The equation becomes:
\( -\frac{1}{3} \left( \frac{1}{v} + \frac{1}{v-1} \right) dv = \frac { 1 }{x} dx \)
Integrate both sides:
\( -\frac{1}{3} \int \left( \frac{1}{v} + \frac{1}{v-1} \right) dv = \int \frac { 1 }{x} dx \)
\( -\frac{1}{3} (\log|v| + \log|v-1|) = \log|x| + C_1 \)
\( -\frac{1}{3} \log|v(v-1)| = \log|x| + C_1 \)
Multiply by -3:
\( \log|v(v-1)| = -3\log|x| - 3C_1 \)
\( \log|v(v-1)| = \log|x^{-3}| + \log|C| \) (where \( -3C_1 = \log|C| \))
\( \log|v(v-1)| = \log\left|\frac{C}{x^3}\right| \)
Remove logarithms:
\( v(v-1) = \frac{C}{x^3} \)
Now, substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x}\left(\frac{y}{x}-1\right) = \frac{C}{x^3} \)
\( \frac{y}{x}\left(\frac{y-x}{x}\right) = \frac{C}{x^3} \)
\( \frac{y(y-x)}{x^2} = \frac{C}{x^3} \)
Multiply by \( x^3 \):
\( xy(y-x) = C \)
This can also be written as \( x(y^2-xy) = C \). This is the general solution for the given differential equation. Note that the sign might change based on the choice of the constant C.
In simple words: We used the substitution 'y = vx' because all parts of the equation had the same power. After separating the 'v' and 'x' terms, we used partial fractions to help with integration. Finally, we replaced 'v' with 'y/x' to get the equation that solves the problem.

🎯 Exam Tip: When using partial fractions, make sure to find the values of A and B correctly. A common mistake is algebraic errors during the substitution or simplification steps.

Question 6. The slope of the tangent to a curve at any point (x, y) on it is given by \((y³ – 2yx²) dx + (2xy² – x³) dy = 0\) and the curve passes through (1, 2). Find the equation of the curve.
Answer:
Given the differential equation: \((y³ – 2yx²) dx + (2xy² – x³) dy = 0\)
This equation describes the slope of the tangent to the curve. We need to find the equation of the curve itself. Rearrange to find \( \frac{dy}{dx} \):
\( (2xy² – x³) dy = -(y³ – 2yx²) dx \)
\( (2xy² – x³) dy = (2yx² – y³) dx \)
\( \frac { dy }{dx} = \frac { 2yx^2-y^3 }{2xy^2-x^3} \) ........ (1)
This is a homogeneous differential equation because all terms in the numerator and denominator are of degree 3.
We use the substitution method. Let \( y = vx \).
Then, \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { 2(vx)x^2-(vx)^3 }{2x(vx)^2-x^3} \)
\( v + x\frac { dv }{dx} = \frac { 2vx^3-v^3x^3 }{2v^2x^3-x^3} \)
\( v + x\frac { dv }{dx} = \frac { x^3(2v-v^3) }{x^3(2v^2-1)} \)
\( v + x\frac { dv }{dx} = \frac { 2v-v^3 }{2v^2-1} \)
Now, separate the variables:
\( x\frac { dv }{dx} = \frac { 2v-v^3 }{2v^2-1} - v \)
\( x\frac { dv }{dx} = \frac { (2v-v^3) - v(2v^2-1) }{2v^2-1} \)
\( x\frac { dv }{dx} = \frac { 2v-v^3 - 2v^3+v }{2v^2-1} \)
\( x\frac { dv }{dx} = \frac { 3v-3v^3 }{2v^2-1} \)
\( x\frac { dv }{dx} = \frac { -3v(v^2-1) }{2v^2-1} \)
Rearrange to integrate:
\( \frac { 2v^2-1 }{3v(v^2-1)} dv = -\frac { 1 }{x} dx \)
For the left side, we use partial fractions: \( \frac { 2v^2-1 }{3v(v-1)(v+1)} = \frac{A}{v} + \frac{B}{v-1} + \frac{C}{v+1} \)
Multiplying by \( 3v(v-1)(v+1) \):
\( 2v^2-1 = 3A(v-1)(v+1) + 3Bv(v+1) + 3Cv(v-1) \)
Set \( v=0 \): \( -1 = 3A(-1)(1) \implies -1 = -3A \implies A = \frac{1}{3} \)
Set \( v=1 \): \( 2(1)^2-1 = 3B(1)(2) \implies 1 = 6B \implies B = \frac{1}{6} \)
Set \( v=-1 \): \( 2(-1)^2-1 = 3C(-1)(-2) \implies 1 = 6C \implies C = \frac{1}{6} \)
So, \( \frac { 2v^2-1 }{3v(v^2-1)} = \frac{1/3}{v} + \frac{1/6}{v-1} + \frac{1/6}{v+1} = \frac{1}{3v} + \frac{1}{6(v-1)} + \frac{1}{6(v+1)} \)
The equation becomes:
\( \left( \frac{1}{3v} + \frac{1}{6(v-1)} + \frac{1}{6(v+1)} \right) dv = -\frac { 1 }{x} dx \)
Integrate both sides:
\( \frac{1}{3}\log|v| + \frac{1}{6}\log|v-1| + \frac{1}{6}\log|v+1| = -\log|x| + C_1 \)
Multiply by 6:
\( 2\log|v| + \log|v-1| + \log|v+1| = -6\log|x| + 6C_1 \)
\( \log|v^2(v-1)(v+1)| = \log|x^{-6}| + \log|K| \) (where \( 6C_1 = \log|K| \))
\( \log|v^2(v^2-1)| = \log\left|\frac{K}{x^6}\right| \)
Remove logarithms:
\( v^2(v^2-1) = \frac{K}{x^6} \)
Now, substitute back \( v = \frac{y}{x} \):
\( \left(\frac{y}{x}\right)^2 \left(\left(\frac{y}{x}\right)^2-1\right) = \frac{K}{x^6} \)
\( \frac{y^2}{x^2} \left(\frac{y^2-x^2}{x^2}\right) = \frac{K}{x^6} \)
\( \frac{y^2(y^2-x^2)}{x^4} = \frac{K}{x^6} \)
Multiply by \( x^6 \):
\( x^2y^2(y^2-x^2) = K \)
This is the general equation of the curve. We need to find the particular solution using the given point (1, 2).
Substitute \( x=1 \) and \( y=2 \) into the equation:
\( (1)^2(2)^2((2)^2-(1)^2) = K \)
\( 1 \cdot 4 (4-1) = K \)
\( 4 \cdot 3 = K \)
\( K = 12 \)
So, the equation of the specific curve passing through (1, 2) is:
\( x^2y^2(y^2-x^2) = 12 \)
In simple words: First, we rearranged the given equation to find the rate of change. Since it was a homogeneous equation, we used 'y = vx' to make it easier to solve. After integrating, we used the given point (1, 2) to find the exact value of the constant, which gave us the final equation of the curve.

🎯 Exam Tip: Always remember to use the given initial conditions (like a point the curve passes through) to find the specific value of the constant of integration, K, for a particular solution.

Question 7. An electric manufacturing company makes small household switches. The company estimates the marginal revenue function for these switches to be \((x² + y²) dy = xy dx\) where x represents the number of units (in thousands). What is the total revenue function?
Answer:
Given the marginal revenue function: \((x² + y²) dy = xy dx\)
To find the total revenue function, we need to solve this differential equation. First, we write it in the standard form \( \frac{dy}{dx} \):
\( \frac { dy }{dx} = \frac { xy }{x^2+y^2} \) ........ (1)
This is a homogeneous differential equation because all terms (x, y) in the numerator and denominator are of degree 2.
We use the substitution method. Let \( y = vx \).
Then, \( \frac { dy }{dx} = v + x\frac { dv }{dx} \).
Substitute these into equation (1):
\( v + x\frac { dv }{dx} = \frac { x(vx) }{x^2+(vx)^2} \)
\( v + x\frac { dv }{dx} = \frac { vx^2 }{x^2+v^2x^2} \)
\( v + x\frac { dv }{dx} = \frac { vx^2 }{x^2(1+v^2)} \)
\( v + x\frac { dv }{dx} = \frac { v }{1+v^2} \)
Now, separate the variables:
\( x\frac { dv }{dx} = \frac { v }{1+v^2} - v \)
\( x\frac { dv }{dx} = \frac { v - v(1+v^2) }{1+v^2} \)
\( x\frac { dv }{dx} = \frac { v - v - v^3 }{1+v^2} \)
\( x\frac { dv }{dx} = \frac { -v^3 }{1+v^2} \)
Rearrange to integrate:
\( \frac { 1+v^2 }{-v^3} dv = \frac { 1 }{x} dx \)
\( -\left( \frac { 1 }{v^3} + \frac { v^2 }{v^3} \right) dv = \frac { 1 }{x} dx \)
\( -\left( v^{-3} + \frac { 1 }{v} \right) dv = \frac { 1 }{x} dx \)
Integrate both sides:
\( - \left( \frac { v^{-2} }{-2} + \log|v| \right) = \log|x| + c_1 \)
\( \frac { 1 }{2v^2} - \log|v| = \log|x| + c_1 \)
Multiply by -1 (to make \(\log|v|\) positive, or simply rearrange):
\( \frac { 1 }{2v^2} = \log|x| + \log|v| + c_1 \)
\( \frac { 1 }{2v^2} = \log|xv| + c_1 \)
Substitute back \( v = \frac{y}{x} \):
\( \frac { 1 }{2\left(\frac{y}{x}\right)^2} = \log\left|x\left(\frac{y}{x}\right)\right| + c_1 \)
\( \frac { 1 }{2\frac{y^2}{x^2}} = \log|y| + c_1 \)
\( \frac { x^2 }{2y^2} = \log|y| + c_1 \)
This is the total revenue function. We can write \( c_1 \) as \( \log|C| \) if preferred, but it's typically left as an additive constant.
Alternatively, we can write \( \frac{x^2}{2y^2} - \log|y| = c_1 \). This represents the total revenue function, where \(c_1\) is a constant determined by initial conditions.
In simple words: We took the equation for marginal revenue, which shows how revenue changes, and used a special trick with 'y=vx' to solve it. After separating and integrating the parts, we put 'y/x' back into the equation. This gave us the total revenue function, which tells us the company's full income.

🎯 Exam Tip: Remember that marginal revenue is the derivative of total revenue. Integrating the marginal revenue function will give the total revenue function, including an arbitrary constant of integration.