Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations More Ques

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems

Question 1. Suppose that \( Q_d = 30 - 5P + 2\frac { dP }{dt} + \frac { d^2P }{dt^2} \) and \( Q_s = 6 + 3P \). Find the equilibrium price for market clearance.
Answer:
We are given the demand function \( Q_d = 30 - 5P + 2\frac { dP }{dt} + \frac { d^2P }{dt^2} \) and the supply function \( Q_s = 6 + 3P \).
For market clearance, the demand must equal the supply: \( Q_d = Q_s \).
So, we set the two equations equal to each other:
\( 30 - 5P + 2\frac { dP }{dt} + \frac { d^2P }{dt^2} = 6 + 3P \)
Now, we rearrange the terms to form a standard differential equation, moving all terms involving P to one side and constants to the other:
\( \frac { d^2P }{dt^2} + 2\frac { dP }{dt} - 5P - 3P + 30 - 6 = 0 \)
\( \frac { d^2P }{dt^2} + 2\frac { dP }{dt} - 8P + 24 = 0 \)
We can write this in operator form, where \( D = \frac{d}{dt} \) and \( D^2 = \frac{d^2}{dt^2} \):
\( (D^2 + 2D - 8) P = -24 \)

First, we find the Complementary Function (C.F.). The auxiliary equation is:
\( m^2 + 2m - 8 = 0 \)
We factor the quadratic equation:
\( (m + 4)(m - 2) = 0 \)
This gives us two distinct real roots:
\( m_1 = -4, m_2 = 2 \)
Since the roots are real and different, the Complementary Function (C.F.) is:
\( C.F. = Ae^{m_1t} + Be^{m_2t} \)
\( C.F. = Ae^{-4t} + Be^{2t} \)

Next, we find the Particular Integral (P.I.). The right-hand side of our differential equation is a constant, \( -24 \).
So, \( P.I. = \frac{1}{D^2 + 2D - 8} (-24) \)
For a constant term, we substitute \( D = 0 \) into the denominator, provided it does not make the denominator zero:
\( P.I. = \frac{1}{(0)^2 + 2(0) - 8} (-24) \)
\( P.I. = \frac{1}{-8} (-24) \)
\( P.I. = 3 \)

The general solution for the equilibrium price \( P \) is the sum of the Complementary Function and the Particular Integral:
\( P = C.F. + P.I. \)
\( P = Ae^{-4t} + Be^{2t} + 3 \)
This equation describes how the price \( P \) changes over time, including constants that depend on initial conditions. Differential equations help model dynamic economic systems.
In simple words: First, we set the demand and supply equations equal to find the market equilibrium. This gives us a special kind of equation called a differential equation. We then solve this equation in two parts: one part shows how the price changes over time by itself, and the other part finds a specific constant price. Putting these two parts together gives the full equation for the price at market clearance.

🎯 Exam Tip: Remember to express the final equilibrium price equation as \( P = C.F. + P.I. \) and clearly show the steps for finding both the Complementary Function and the Particular Integral.

Question 2. Form the differential equation having for its general solution \( y = ax^2 + bx \)
Answer:
We are given the general solution \( y = ax^2 + bx \). This equation has two arbitrary constants, \( a \) and \( b \). To form a differential equation from it, we need to differentiate twice, which will help us eliminate these two constants.

Given equation:
\( y = ax^2 + bx \) ....... (1)

First differentiation with respect to \( x \):
\( \frac{dy}{dx} = 2ax + b \) ....... (2)

Second differentiation with respect to \( x \):
\( \frac{d^2y}{dx^2} = 2a \) ....... (3)

From equation (3), we can find the value of \( a \):
\( a = \frac{1}{2} \frac{d^2y}{dx^2} \)

Now, substitute the value of \( a \) back into equation (2) to find \( b \):
\( \frac{dy}{dx} = 2 \left( \frac{1}{2} \frac{d^2y}{dx^2} \right) x + b \)
\( \frac{dy}{dx} = x \frac{d^2y}{dx^2} + b \)

So, the value of \( b \) is:
\( b = \frac{dy}{dx} - x \frac{d^2y}{dx^2} \) ....... (4)

Finally, substitute the values of \( a \) and \( b \) from equation (3) and (4) into the original equation (1):
\( y = \left( \frac{1}{2} \frac{d^2y}{dx^2} \right) x^2 + \left( \frac{dy}{dx} - x \frac{d^2y}{dx^2} \right) x \)
\( y = \frac{x^2}{2} \frac{d^2y}{dx^2} + x \frac{dy}{dx} - x^2 \frac{d^2y}{dx^2} \)
To simplify, we can multiply the entire equation by 2 to clear the fraction:
\( 2y = x^2 \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} - 2x^2 \frac{d^2y}{dx^2} \)
Combine the terms involving \( \frac{d^2y}{dx^2} \):
\( 2y = (x^2 - 2x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} \)
\( 2y = -x^2 \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} \)
Rearrange the terms to get the standard form of a differential equation:
\( x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 \)
This is the required differential equation. Forming a differential equation from a general solution often involves successive differentiation to eliminate arbitrary constants.
In simple words: We start with the given solution and differentiate it two times because it has two unknown constants. Each time we differentiate, we get a new equation. Then, we use these new equations to find what the constants are in terms of y, dy/dx, and d²y/dx². Finally, we put these back into the first equation to get rid of the constants and form the differential equation.

🎯 Exam Tip: The number of arbitrary constants in the general solution tells you the order of the differential equation you need to form. Here, two constants mean a second-order differential equation.

Question 3. Solve \( yx^2dx + e^{-x} dy = 0 \)
Answer:
We need to solve the given differential equation: \( yx^2dx + e^{-x} dy = 0 \).
This is a separable differential equation, meaning we can separate the variables \( x \) and \( y \) to opposite sides.

First, move the \( yx^2dx \) term to the right side:
\( e^{-x} dy = -yx^2dx \)
Now, divide by \( e^{-x} \) and \( y \) to separate the variables:
\( \frac{1}{y} dy = \frac{-x^2}{e^{-x}} dx \)
We can rewrite \( \frac{1}{e^{-x}} \) as \( e^x \):
\( \frac{1}{y} dy = -x^2e^x dx \)

Now, integrate both sides:
\( \int \frac{1}{y} dy = \int -x^2e^x dx \)
\( \log|y| = - \int x^2e^x dx \)
To solve the integral \( \int x^2e^x dx \), we use integration by parts, which has the formula: \( \int u dv = uv - \int v du \). We can apply this twice or use a generalized formula for repeated integration by parts, often called tabular integration:
\( \int u dv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \)
Here, let \( u = x^2 \) and \( dv = e^x dx \).
Then \( u' = 2x \), \( u'' = 2 \), \( u''' = 0 \).
And \( v = \int e^x dx = e^x \), \( v_1 = \int e^x dx = e^x \), \( v_2 = \int e^x dx = e^x \).
So, \( \int x^2e^x dx = x^2e^x - (2x)e^x + (2)e^x \)
\( \int x^2e^x dx = e^x(x^2 - 2x + 2) \)

Substitute this back into our main equation:
\( \log|y| = - [e^x(x^2 - 2x + 2)] + C \)
We can also write the constant as \( \log C \) for convenience:
\( \log|y| = -e^x(x^2 - 2x + 2) + \log C \)
Rearrange the terms:
\( \log|y| - \log C = -e^x(x^2 - 2x + 2) \)
\( \log \left| \frac{y}{C} \right| = -e^x(x^2 - 2x + 2) \)
To remove the logarithm, we use the exponential function:
\( \frac{y}{C} = e^{-e^x(x^2 - 2x + 2)} \)
\( y = Ce^{-e^x(x^2 - 2x + 2)} \)
This is the general solution to the differential equation. The key step is separating variables and then carefully applying integration by parts.
In simple words: We first move all the 'y' terms to one side with 'dy' and all the 'x' terms to the other side with 'dx'. Then, we integrate both sides. For the 'x' side, we use a special method called integration by parts because we have 'x²' multiplied by 'e^x'. After integrating, we solve for 'y' to get the final answer.

🎯 Exam Tip: For separable differential equations, always aim to gather all terms involving one variable with its differential on one side, and the other variable with its differential on the other side, before integrating.

Question 4. Solve \( (x^2 + y^2) dx + 2xy dy = 0 \)
Answer:
We need to solve the differential equation: \( (x^2 + y^2) dx + 2xy dy = 0 \).
First, let's rearrange it to the form \( \frac{dy}{dx} = f(x, y) \):
\( 2xy dy = -(x^2 + y^2) dx \)
\( \frac{dy}{dx} = \frac{-(x^2 + y^2)}{2xy} \) ....... (1)

Now, we check if this is a homogeneous differential equation. A function \( f(x,y) \) is homogeneous of degree \( n \) if \( f(kx, ky) = k^n f(x,y) \). Here, if we replace \( x \) with \( kx \) and \( y \) with \( ky \):
\( \frac{-((kx)^2 + (ky)^2)}{2(kx)(ky)} = \frac{-(k^2x^2 + k^2y^2)}{2k^2xy} = \frac{-k^2(x^2 + y^2)}{k^2(2xy)} = \frac{-(x^2 + y^2)}{2xy} \)
Since the \( k \) terms cancel out, the function is homogeneous of degree 0. So, this is a homogeneous differential equation.

For homogeneous equations, we use the substitution \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{-(x^2 + (vx)^2)}{2x(vx)} \)
\( v + x \frac{dv}{dx} = \frac{-(x^2 + v^2x^2)}{2vx^2} \)
Factor out \( x^2 \) from the numerator:
\( v + x \frac{dv}{dx} = \frac{-x^2(1 + v^2)}{2vx^2} \)
Cancel \( x^2 \) from numerator and denominator:
\( v + x \frac{dv}{dx} = \frac{-(1 + v^2)}{2v} \)
Now, move \( v \) to the right side:
\( x \frac{dv}{dx} = \frac{-(1 + v^2)}{2v} - v \)
Combine the terms on the right side:
\( x \frac{dv}{dx} = \frac{-(1 + v^2) - 2v^2}{2v} \)
\( x \frac{dv}{dx} = \frac{-1 - v^2 - 2v^2}{2v} \)
\( x \frac{dv}{dx} = \frac{-1 - 3v^2}{2v} \)
\( x \frac{dv}{dx} = -\frac{1 + 3v^2}{2v} \)

Now, separate the variables \( v \) and \( x \):
\( \frac{2v}{1 + 3v^2} dv = -\frac{1}{x} dx \)

Integrate both sides:
\( \int \frac{2v}{1 + 3v^2} dv = \int -\frac{1}{x} dx \)
For the left integral, let \( u = 1 + 3v^2 \), then \( du = 6v dv \), so \( 2v dv = \frac{1}{3} du \).
\( \int \frac{1}{u} \left( \frac{1}{3} du \right) = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} \log|u| \)
Substituting back \( u = 1 + 3v^2 \):
\( \frac{1}{3} \log|1 + 3v^2| \)
For the right integral:
\( \int -\frac{1}{x} dx = -\log|x| + \log C \)
So, we have:
\( \frac{1}{3} \log|1 + 3v^2| = -\log|x| + \log C \)
Multiply by 3:
\( \log|1 + 3v^2| = -3\log|x| + 3\log C \)
\( \log|1 + 3v^2| = \log|x^{-3}| + \log C^3 \)
Let \( K = C^3 \) (another constant):
\( \log|1 + 3v^2| = \log \left| \frac{K}{x^3} \right| \)
Remove the logarithms:
\( 1 + 3v^2 = \frac{K}{x^3} \)
Now, substitute back \( v = \frac{y}{x} \):
\( 1 + 3\left(\frac{y}{x}\right)^2 = \frac{K}{x^3} \)
\( 1 + \frac{3y^2}{x^2} = \frac{K}{x^3} \)
Multiply by \( x^2 \) to clear the fraction on the left:
\( x^2 + 3y^2 = \frac{K}{x} \)
Multiply by \( x \):
\( x(x^2 + 3y^2) = K \)
\( x^3 + 3xy^2 = K \)
This is the general solution. Homogeneous differential equations can often be simplified using the substitution \( y = vx \).
In simple words: First, we rewrite the equation to find dy/dx. We see that all terms in the equation have the same total power of x and y, which means it is a "homogeneous" equation. To solve this, we replace 'y' with 'vx' and 'dy/dx' with 'v + x dv/dx'. After simplifying, we separate the 'v' terms and 'x' terms and integrate both sides. Finally, we put 'y/x' back in place of 'v' to get the answer.

🎯 Exam Tip: Always verify if the given differential equation is homogeneous before applying the substitution \( y=vx \). This involves checking the degree of all terms in \( M(x,y) \) and \( N(x,y) \) or checking \( f(kx,ky)=f(x,y) \).

Question 5. Solve \( x \frac{dy}{dx} + 2y = x^4 \)
Answer:
We are asked to solve the differential equation: \( x \frac{dy}{dx} + 2y = x^4 \).
This is a first-order linear differential equation, which has the general form: \( \frac{dy}{dx} + P(x)y = Q(x) \).

First, we need to divide the entire equation by \( x \) to get it into the standard linear form:
\( \frac{dy}{dx} + \frac{2}{x}y = x^3 \)

Now, we can identify \( P(x) = \frac{2}{x} \) and \( Q(x) = x^3 \).
Next, we calculate the Integrating Factor (I.F.), which is given by \( e^{\int P(x) dx} \).
\( \int P(x) dx = \int \frac{2}{x} dx = 2 \int \frac{1}{x} dx = 2 \log|x| = \log(x^2) \)
So, the Integrating Factor (I.F.) is:
\( I.F. = e^{\log(x^2)} = x^2 \)

The general solution for a first-order linear differential equation is given by:
\( y \times (I.F.) = \int Q(x) \times (I.F.) dx + C \)
Substitute the values:
\( y \times x^2 = \int x^3 \times x^2 dx + C \)
\( yx^2 = \int x^5 dx + C \)
Now, integrate \( x^5 \):
\( yx^2 = \frac{x^{5+1}}{5+1} + C \)
\( yx^2 = \frac{x^6}{6} + C \)
Finally, we can solve for \( y \):
\( y = \frac{x^6}{6x^2} + \frac{C}{x^2} \)
\( y = \frac{x^4}{6} + \frac{C}{x^2} \)
This is the general solution to the given differential equation. Recognizing the type of differential equation is the first step to solving it correctly.
In simple words: First, we change the equation into a special form for "linear differential equations". Then, we find something called an "Integrating Factor" by using a small calculation with \( P(x) \). We multiply this factor with the whole equation. After that, we integrate both sides and solve for \( y \) to find the answer.

🎯 Exam Tip: Always ensure the coefficient of \( \frac{dy}{dx} \) is 1 before identifying \( P(x) \) and \( Q(x) \) for calculating the Integrating Factor.

Question 6. A manufacturing company has found that the cost \( C \) of operating and maintaining the equipment is related to the length \( 'm' \) of intervals between overhauls by the equation \( m^2 \frac { dC }{dm} + 2mC = 2 \) and \( C = 4 \) when \( m = 2 \). Find the relationship between \( C \) and \( m \).
Answer:
We are given the differential equation relating cost \( C \) and interval length \( m \):
\( m^2 \frac { dC }{dm} + 2mC = 2 \)
This is a first-order linear differential equation. To solve it, we first divide by \( m^2 \) to get it into the standard form \( \frac{dC}{dm} + P(m)C = Q(m) \):
\( \frac { dC }{dm} + \frac { 2m }{m^2}C = \frac { 2 }{m^2} \)
\( \frac { dC }{dm} + \frac { 2 }{m}C = \frac { 2 }{m^2} \)

Now we identify \( P(m) = \frac{2}{m} \) and \( Q(m) = \frac{2}{m^2} \).
Next, we calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P(m) dm} \)
\( \int P(m) dm = \int \frac{2}{m} dm = 2 \log|m| = \log(m^2) \)
So, \( I.F. = e^{\log(m^2)} = m^2 \)

The general solution is given by:
\( C \times (I.F.) = \int Q(m) \times (I.F.) dm + k \)
\( C m^2 = \int \frac { 2 }{m^2} \times m^2 dm + k \)
\( C m^2 = \int 2 dm + k \)
\( C m^2 = 2m + k \) ....... (1)

We are given the initial condition that \( C = 4 \) when \( m = 2 \). We use this to find the constant \( k \):
\( (4)(2)^2 = 2(2) + k \)
\( 4 \times 4 = 4 + k \)
\( 16 = 4 + k \)
\( k = 16 - 4 \)
\( k = 12 \)

Substitute the value of \( k \) back into equation (1) to get the specific relationship between \( C \) and \( m \):
\( C m^2 = 2m + 12 \)
We can also express \( C \) in terms of \( m \):
\( C = \frac{2m + 12}{m^2} \)
\( C = \frac{2(m + 6)}{m^2} \)
This equation describes the cost \( C \) as a function of the interval length \( m \). Understanding how costs change based on maintenance schedules is crucial for business decisions.
In simple words: We start with the given equation that links cost \( C \) and maintenance interval \( m \). We change it into a standard form to solve. Then, we find an "Integrating Factor" and use it to solve the equation, which gives us a general answer with an unknown constant. We use the given information (C=4 when m=2) to find this constant. Finally, we write down the full relationship between \( C \) and \( m \).

🎯 Exam Tip: When given initial conditions, remember to use them to find the specific value of the constant of integration, making the solution particular rather than general.

Question 7. Solve \( (D^2 - 3D + 2) y = e^{4x} \)
Answer:
We need to solve the second-order linear differential equation: \( (D^2 - 3D + 2) y = e^{4x} \).

First, we find the Complementary Function (C.F.) by solving the auxiliary equation:
\( m^2 - 3m + 2 = 0 \)
Factor the quadratic equation:
\( (m - 1)(m - 2) = 0 \)
This gives us two distinct real roots:
\( m_1 = 1, m_2 = 2 \)
Since the roots are real and different, the Complementary Function (C.F.) is:
\( C.F. = Ae^{m_1x} + Be^{m_2x} \)
\( C.F. = Ae^x + Be^{2x} \)

Next, we find the Particular Integral (P.I.). The right-hand side is \( e^{4x} \).
\( P.I. = \frac{1}{D^2 - 3D + 2} e^{4x} \)
For \( e^{ax} \) type, we substitute \( D = a \) (here \( a = 4 \)) into the denominator, provided it does not make the denominator zero.
Let's check the denominator with \( D = 4 \):
\( (4)^2 - 3(4) + 2 = 16 - 12 + 2 = 6 \)
Since the denominator is not zero, we can directly substitute \( D = 4 \):
\( P.I. = \frac{1}{(4)^2 - 3(4) + 2} e^{4x} \)
\( P.I. = \frac{1}{16 - 12 + 2} e^{4x} \)
\( P.I. = \frac{1}{6} e^{4x} \)

The general solution is the sum of the Complementary Function and the Particular Integral:
\( y = C.F. + P.I. \)
\( y = Ae^x + Be^{2x} + \frac{e^{4x}}{6} \)
This is the general solution for \( y \). Solving these equations helps understand systems that change over time based on specific driving forces.
In simple words: To solve this equation, we first find the "Complementary Function" by solving a simple quadratic equation that comes from the D-terms. Then, we find the "Particular Integral" by replacing 'D' with the number from the power of 'e' on the right side. If that doesn't make the bottom zero, we use it directly. Finally, we add these two parts together to get the full solution for \( y \).

🎯 Exam Tip: Always check if the value you substitute for D in the P.I. (for \( e^{ax} \) forms) makes the denominator zero. If it does, you need to use the special rule (multiply by x and differentiate the denominator).

Question 8. Solve \( \frac{dy}{dx} + y \cos x = 2 \cos x \)
Answer:
We need to solve the differential equation: \( \frac{dy}{dx} + y \cos x = 2 \cos x \).
This is a first-order linear differential equation in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).

Here, we can identify \( P(x) = \cos x \) and \( Q(x) = 2 \cos x \).
Next, we calculate the Integrating Factor (I.F.):
\( I.F. = e^{\int P(x) dx} \)
\( \int P(x) dx = \int \cos x dx = \sin x \)
So, the Integrating Factor (I.F.) is:
\( I.F. = e^{\sin x} \)

The general solution is given by:
\( y \times (I.F.) = \int Q(x) \times (I.F.) dx + C \)
\( y e^{\sin x} = \int (2 \cos x) e^{\sin x} dx + C \)
To solve the integral on the right, let \( t = \sin x \). Then \( dt = \cos x dx \).
So, the integral becomes:
\( \int (2 \cos x) e^{\sin x} dx = \int 2 e^t dt \)
\( = 2 \int e^t dt = 2e^t \)
Substitute back \( t = \sin x \):
\( = 2e^{\sin x} \)

Now, substitute this back into the general solution formula:
\( y e^{\sin x} = 2e^{\sin x} + C \)
Finally, divide by \( e^{\sin x} \) to solve for \( y \):
\( y = \frac{2e^{\sin x}}{e^{\sin x}} + \frac{C}{e^{\sin x}} \)
\( y = 2 + Ce^{-\sin x} \)
This is the general solution. Using an integrating factor is a standard technique for these types of equations.
In simple words: We recognize this as a "linear" differential equation. We find something called an "Integrating Factor" which is \( e \) raised to the power of the integral of \( \cos x \). Then, we use a special formula: \( y \) times the Integrating Factor equals the integral of \( Q(x) \) times the Integrating Factor. We solve this integral and then find \( y \).

🎯 Exam Tip: When using the substitution method for the integral \( \int Q(x) \times (I.F.) dx \), clearly define your substitution (e.g., \( t = \sin x \)) and its derivative (e.g., \( dt = \cos x dx \)) to avoid errors.

Question 9. Solve \( x^2y dx - (x^3 + y^3) dy = 0 \)
Answer:
We need to solve the differential equation: \( x^2y dx - (x^3 + y^3) dy = 0 \).
First, we rearrange it to the form \( \frac{dy}{dx} = f(x, y) \):
\( (x^3 + y^3) dy = x^2y dx \)
\( \frac{dy}{dx} = \frac{x^2y}{x^3 + y^3} \) ....... (1)

Now, we check if this is a homogeneous differential equation. A function is homogeneous if \( f(kx, ky) = k^n f(x,y) \). Here:
\( \frac{(kx)^2(ky)}{(kx)^3 + (ky)^3} = \frac{k^2x^2ky}{k^3x^3 + k^3y^3} = \frac{k^3x^2y}{k^3(x^3 + y^3)} = \frac{x^2y}{x^3 + y^3} \)
Since the \( k \) terms cancel out, this is a homogeneous differential equation of degree 0.

For homogeneous equations, we use the substitution \( y = vx \).
Then, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{x^2(vx)}{x^3 + (vx)^3} \)
\( v + x \frac{dv}{dx} = \frac{vx^3}{x^3 + v^3x^3} \)
Factor out \( x^3 \) from the denominator:
\( v + x \frac{dv}{dx} = \frac{vx^3}{x^3(1 + v^3)} \)
Cancel \( x^3 \):
\( v + x \frac{dv}{dx} = \frac{v}{1 + v^3} \)
Now, move \( v \) to the right side:
\( x \frac{dv}{dx} = \frac{v}{1 + v^3} - v \)
Combine the terms on the right side:
\( x \frac{dv}{dx} = \frac{v - v(1 + v^3)}{1 + v^3} \)
\( x \frac{dv}{dx} = \frac{v - v - v^4}{1 + v^3} \)
\( x \frac{dv}{dx} = \frac{-v^4}{1 + v^3} \)

Now, separate the variables \( v \) and \( x \):
\( \frac{1 + v^3}{-v^4} dv = \frac{1}{x} dx \)
\( -\frac{1 + v^3}{v^4} dv = \frac{1}{x} dx \)
\( -\left( \frac{1}{v^4} + \frac{v^3}{v^4} \right) dv = \frac{1}{x} dx \)
\( -\left( v^{-4} + \frac{1}{v} \right) dv = \frac{1}{x} dx \)

Integrate both sides:
\( \int -\left( v^{-4} + \frac{1}{v} \right) dv = \int \frac{1}{x} dx \)
\( -\left( \frac{v^{-3}}{-3} + \log|v| \right) = \log|x| + C \)
\( \frac{v^{-3}}{3} - \log|v| = \log|x| + C \)
\( \frac{1}{3v^3} - \log|v| = \log|x| + C \)
\( \frac{1}{3v^3} = \log|x| + \log|v| + C \)
\( \frac{1}{3v^3} = \log|vx| + C \)
Now, substitute back \( v = \frac{y}{x} \):
\( \frac{1}{3(\frac{y}{x})^3} = \log\left| \frac{y}{x}x \right| + C \)
\( \frac{x^3}{3y^3} = \log|y| + C \)
Rearrange the terms:
\( \log|y| = \frac{x^3}{3y^3} - C \)
This is the general solution. Homogeneous equations are simplified by reducing them to separable form using substitution.
In simple words: First, we rearrange the equation to find dy/dx. We notice that all terms have the same total power of x and y, which makes it a "homogeneous" equation. We replace 'y' with 'vx' and 'dy/dx' with 'v + x dv/dx'. After doing some algebra, we separate the 'v' terms to one side and 'x' terms to the other. Then, we integrate both sides. Finally, we put 'y/x' back in place of 'v' and simplify to get the answer.

🎯 Exam Tip: When integrating \( -\left( v^{-4} + \frac{1}{v} \right) dv \), remember that \( \int v^{-4} dv = \frac{v^{-3}}{-3} \) and \( \int \frac{1}{v} dv = \log|v| \). Don't confuse the power rule with the logarithm rule.

Question 10. Solve \( \frac{dy}{dx} = xy + x + y + 1 \)
Answer:
We need to solve the differential equation: \( \frac{dy}{dx} = xy + x + y + 1 \).
This equation can be solved by separating variables. First, we group terms on the right-hand side:
\( \frac{dy}{dx} = (xy + x) + (y + 1) \)
Factor out \( x \) from the first group:
\( \frac{dy}{dx} = x(y + 1) + (y + 1) \)
Now, factor out the common term \( (y + 1) \):
\( \frac{dy}{dx} = (x + 1)(y + 1) \)

Now that the variables are separated, we can move all \( y \) terms to the left side with \( dy \) and all \( x \) terms to the right side with \( dx \):
\( \frac{1}{y + 1} dy = (x + 1) dx \)

Now, integrate both sides:
\( \int \frac{1}{y + 1} dy = \int (x + 1) dx \)
For the left side, the integral of \( \frac{1}{u} \) is \( \log|u| \):
\( \log|y + 1| \)
For the right side, we integrate term by term:
\( \int x dx + \int 1 dx = \frac{x^2}{2} + x \)
So, we have:
\( \log|y + 1| = \frac{x^2}{2} + x + C \)
This is the general solution for the differential equation. Separating variables is a fundamental technique for many first-order differential equations.
In simple words: First, we rewrite the equation by grouping the terms on the right side and factoring them. This helps us separate the 'y' terms with 'dy' on one side and the 'x' terms with 'dx' on the other. Then, we integrate both sides. After integrating, we get the final equation that links \( y \) and \( x \).

🎯 Exam Tip: Always look for ways to factor the right-hand side of a first-order differential equation to achieve separation of variables, as this often simplifies the integration process significantly.