Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations Exercise 4.2

Question 1. Solve:
(i) \( \frac { dy }{dx} = ae^y \)
Answer:
Given the differential equation: \( \frac { dy }{dx} = ae^y \)
We rearrange the terms to separate the variables:
\( \frac { dy }{e^y} = adx \)
This can be written as:
\( e^{-y} dy = adx \)
Next, we integrate both sides of the equation:
\( \int e^{-y} dy = \int adx \)
\( \frac { e^{-y} }{(-1)} = ax + c \)
\( -e^{-y} = ax + c \)

\( \implies e^{-y} + ax + c = 0 \)
This is the general solution for the given differential equation. The constant 'a' and 'c' are arbitrary constants determined by initial conditions.
In simple words: We separate the 'y' terms with 'dy' and 'x' terms with 'dx'. Then we use integration to solve for 'y' and 'x', ending up with a single equation that includes a constant.

🎯 Exam Tip: Always remember to add the constant of integration 'c' when solving indefinite integrals. If initial conditions are given, use them to find the value of 'c'.

Question 1. (ii) \( \frac { 1+x^2 }{1+y} = xy \frac { dy }{dx} \)
Answer:
Given the differential equation:
\( \frac { 1+x^2 }{1+y} = xy \frac { dy }{dx} \)
First, we separate the variables 'x' and 'y':
\( \frac { 1+x^2 }{x} \cdot \frac{1}{x} dx = y(1+y) dy \)
This simplifies to:
\( (\frac { 1 }{x} + x) dx = (y + y^2) dy \)
Now, we integrate both sides:
\( \int (\frac { 1 }{x} + x) dx = \int (y + y^2) dy \)
Integrating term by term, we get:
\( \log |x| + \frac { x^2 }{2} = \frac { y^2 }{2} + \frac { y^3 }{3} + c \)
This is the general solution for the given differential equation. Separating variables is a key technique here.
In simple words: We put all the 'x' parts with 'dx' on one side and all the 'y' parts with 'dy' on the other. Then we do the sum backwards (integrate) on both sides to find the answer, adding a constant at the end.

🎯 Exam Tip: When separating variables, be careful with multiplication and division to ensure all x-terms are with dx and all y-terms are with dy.

Question 2. \( y(1 - x) - x \frac { dy }{dx} = 0 \)
Answer:
Given the differential equation:
\( y(1 - x) - x \frac { dy }{dx} = 0 \)
First, we isolate the \( \frac { dy }{dx} \) term:
\( y(1 - x) = x \frac { dy }{dx} \)
Now, separate the variables 'x' and 'y':
\( \frac { (1 - x) }{x} dx = \frac { 1 }{y} dy \)
This can be written as:
\( (\frac { 1 }{x} - 1) dx = \frac { 1 }{y} dy \)
Next, integrate both sides:
\( \int (\frac { 1 }{x} - 1) dx = \int \frac { 1 }{y} dy \)
Integrating gives us:
\( \log |x| - x = \log |y| + c \)
This is the general solution. Always remember that the absolute value is used with logarithms for proper domain.
In simple words: We move all the 'x' parts to one side with 'dx' and all the 'y' parts to the other side with 'dy'. Then we integrate both sides to find the solution.

🎯 Exam Tip: When integrating \( \frac{1}{x} \), the result is \( \log |x| \), which is important to remember for proper function definition.

Question 3.
(i) \( ydx - xdy = 0 \)
Answer:
Given the differential equation:
\( ydx - xdy = 0 \)
First, we separate the variables 'x' and 'y':
\( ydx = xdy \)
Divide both sides by \( xy \):
\( \frac { 1 }{x} dx = \frac { 1 }{y} dy \)
Now, we integrate both sides:
\( \int \frac { 1 }{x} dx = \int \frac { 1 }{y} dy \)
Integrating gives us:
\( \log |x| = \log |y| + \log |c| \)
Using logarithm properties, \( \log |y| + \log |c| = \log |cy| \):
\( \log |x| = \log |cy| \)

\( \implies x = cy \)
This is the general solution. The constant 'c' is an arbitrary constant.
In simple words: We put all the 'x' parts with 'dx' on one side and all the 'y' parts with 'dy' on the other. Then, we integrate both sides and use logarithm rules to get the final answer.

🎯 Exam Tip: When all terms in the integration result in logarithms, it's often simpler to express the constant of integration as \( \log c \) to combine terms neatly.

Question 3. (ii) \( \frac { dy }{dx} + e^x + ye^x = 0 \)
Answer:
Given the differential equation:
\( \frac { dy }{dx} + e^x + ye^x = 0 \)
First, we factor out \( e^x \) from the last two terms:
\( \frac { dy }{dx} + e^x(1 + y) = 0 \)
Now, rearrange the terms to separate variables:
\( \frac { dy }{dx} = -e^x(1 + y) \)
\( \frac { dy }{(1 + y)} = -e^x dx \)
Next, integrate both sides:
\( \int \frac { 1 }{(1 + y)} dy = \int -e^x dx \)
Integrating gives us:
\( \log |1 + y| = -e^x + c \)
This is the general solution for the given differential equation. Factoring is crucial for separation.
In simple words: We first take out the common \( e^x \) part. Then we move all the 'y' parts with 'dy' to one side and all the 'x' parts with 'dx' to the other. After that, we do the integration on both sides to find the answer.

🎯 Exam Tip: Always look for common factors to simplify expressions before attempting to separate variables; this often makes the integration steps much easier.

Question 4. Solve : \( \cos x (1 + \cos y) dx - \sin y (1 + \sin x) dy = 0 \)
Answer:
Given the differential equation:
\( \cos x (1 + \cos y) dx - \sin y (1 + \sin x) dy = 0 \)
First, move the second term to the right side:
\( \cos x (1 + \cos y) dx = \sin y (1 + \sin x) dy \)
Now, separate the variables 'x' and 'y':
\( \frac { \cos x }{(1 + \sin x)} dx = \frac { \sin y }{(1 + \cos y)} dy \)
Next, integrate both sides:
\( \int \frac { \cos x }{(1 + \sin x)} dx = \int \frac { \sin y }{(1 + \cos y)} dy \)
For the left side, let \( u = 1 + \sin x \), so \( du = \cos x dx \). For the right side, let \( v = 1 + \cos y \), so \( dv = -\sin y dy \).
So, the integral becomes:
\( \int \frac { 1 }{u} du = \int -\frac { 1 }{v} dv \)
\( \log |1 + \sin x| = -\log |1 + \cos y| + \log |c| \)
Move the negative log term to the left side:
\( \log |1 + \sin x| + \log |1 + \cos y| = \log |c| \)
Using logarithm properties, \( \log A + \log B = \log (AB) \):
\( \log |(1 + \sin x)(1 + \cos y)| = \log |c| \)

\( \implies (1 + \sin x)(1 + \cos y) = c \)
This is the general solution. Using substitution makes these integrals straightforward.
In simple words: We first put all the 'x' terms with 'dx' on one side and all the 'y' terms with 'dy' on the other. Then we integrate both sides. We use a trick with logarithms where adding logs means multiplying their insides to get a simpler final answer.

🎯 Exam Tip: For integrals of the form \( \int \frac{f'(x)}{f(x)} dx \), the result is \( \log |f(x)| \). Recognize this pattern to quickly solve such integrals.

Question 5. Solve: \( (1-x) dy - (1 + y) dx = 0 \)
Answer:
Given the differential equation:
\( (1-x) dy - (1 + y) dx = 0 \)
First, move the negative term to the right side:
\( (1-x) dy = (1 + y) dx \)
Now, separate the variables 'x' and 'y':
\( \frac { dy }{(1 + y)} = \frac { dx }{(1 - x)} \)
Next, integrate both sides:
\( \int \frac { dy }{(1 + y)} = \int \frac { dx }{(1 - x)} \)
Integrating gives us:
\( \log |1 + y| = -\log |1 - x| + \log |c| \)
Move the negative log term to the left side:
\( \log |1 + y| + \log |1 - x| = \log |c| \)
Using logarithm properties, \( \log A + \log B = \log (AB) \):
\( \log |(1 + y)(1 - x)| = \log |c| \)

\( \implies (1 - x)(1 + y) = c \)
This is the general solution. Combining constants under a single logarithm helps simplify the result.
In simple words: We move all the 'x' parts to one side with 'dx' and all the 'y' parts to the other side with 'dy'. Then we integrate both sides and use logarithm rules to combine them into one simple equation.

🎯 Exam Tip: Be careful with the sign when integrating \( \frac{1}{a-x} \), as it results in \( -\log|a-x| \).

Question 6. Solve:
(i) \( \frac { dy }{dx} = y \sin 2x \)
Answer:
Given the differential equation:
\( \frac { dy }{dx} = y \sin 2x \)
First, separate the variables 'x' and 'y':
\( \frac { dy }{y} = \sin 2x dx \)
Next, integrate both sides:
\( \int \frac { 1 }{y} dy = \int \sin 2x dx \)
Integrating gives us:
\( \log |y| = -\frac { \cos 2x }{2} + c \)
This is the general solution. Integrating \( \sin(ax) \) gives \( -\frac{1}{a}\cos(ax) \).
In simple words: We put all the 'y' parts with 'dy' on one side and all the 'x' parts with 'dx' on the other. Then we perform integration on both sides to find the answer.

🎯 Exam Tip: Remember the integration rule for trigonometric functions like \( \int \sin(ax) dx = -\frac{1}{a} \cos(ax) + C \).

Question 6. (ii) \( \log(\frac { dy }{dx}) = ax + by \)
Answer:
Given the differential equation:
\( \log(\frac { dy }{dx}) = ax + by \)
To remove the logarithm, we convert to exponential form:
\( \frac { dy }{dx} = e^{(ax + by)} \)
Using exponent rules, this can be split:
\( \frac { dy }{dx} = e^{ax} \cdot e^{by} \)
Now, separate the variables 'x' and 'y':
\( \frac { dy }{e^{by}} = e^{ax} dx \)
This can be written as:
\( e^{-by} dy = e^{ax} dx \)
Next, integrate both sides:
\( \int e^{-by} dy = \int e^{ax} dx \)
Integrating gives us:
\( \frac { e^{-by} }{(-b)} = \frac { e^{ax} }{a} + c \)
This is the general solution. Exponential functions are often solved by separating variables.
In simple words: First, we change the log equation into an exponential one. Then, we move all the 'y' parts with 'dy' to one side and all the 'x' parts with 'dx' to the other. Finally, we integrate both sides to get the solution.

🎯 Exam Tip: When \( \log X = Y \), it means \( X = e^Y \). This is a crucial step for solving differential equations involving logarithms.

Question 7. Find the curve whose gradient at any point P (x, y) on it is \( \frac {x-a }{y-b} \) and which passes through the origin.
Answer:
The gradient of a curve at a point (x, y) is given by \( \frac { dy }{dx} \).
So, we have the differential equation:
\( \frac { dy }{dx} = \frac {x-a }{y-b} \)
First, separate the variables 'x' and 'y':
\( (y - b) dy = (x - a) dx \)
Next, integrate both sides:
\( \int (y - b) dy = \int (x - a) dx \)
Integrating gives us:
\( \frac { (y - b)^2 }{2} = \frac { (x - a)^2 }{2} + c \)
Multiply the entire equation by 2:
\( (y - b)^2 = (x - a)^2 + 2c \)
Let \( 2c = C \), where \( C \) is a new constant:
\( (y - b)^2 = (x - a)^2 + C \) ... (1)
We are given that the curve passes through the origin (0, 0). Substitute \( x = 0 \) and \( y = 0 \) into equation (1):
\( (0 - b)^2 = (0 - a)^2 + C \)
\( b^2 = a^2 + C \)

\( \implies C = b^2 - a^2 \) ... (2)
Now, substitute the value of \( C \) from (2) back into equation (1):
\( (y - b)^2 = (x - a)^2 + b^2 - a^2 \)
This is the equation of the curve. The initial condition helps find the specific constant.
In simple words: We are given how the slope of a curve changes. We separate the 'y' and 'x' parts and integrate them. This gives us a general equation with a constant. Since the curve goes through the point (0,0), we use this information to find what that constant should be, giving us the exact equation for the curve.

🎯 Exam Tip: When solving for a specific curve, always use the given initial conditions (like passing through the origin) to determine the exact value of the constant of integration.