Homogeneous Differential Equations

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Homogeneous Differential Equations

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Points to Remember

A homogeneous differential equation is a first-order differential equation of the form

$\frac{dy}{dx} = F\left( \frac{y}{x} \right)$ where the right-hand side is a function of $y/x$ only.

Such equations are reducible to separable form by the substitution:

$y = ux \quad \Rightarrow \quad \frac{dy}{dx} = u + x \frac{du}{dx}$

General Steps to Solve a Homogeneous Differential Equation

Step 1. Confirm that the equation is homogeneous:

Check if the right-hand side of the equation is a function of $y/x$ or $x/y$ only.

Step 2. Use the substitution: $y = ux$ or $u = \frac{y}{x}$

This gives: $\frac{dy}{dx} = u + x \frac{du}{dx}$

Step 3. Substitute into the original differential equation:

Replace all $y$ with $ux$, and $\frac{dy}{dx}$ with $u + x \frac{du}{dx}$.

Step 4. Simplify and rearrange to isolate $\frac{du}{dx}$.

Step 5. Separate the variables:

Rearrange into the form:

$\text{function of } u ; du = \text{function of } x ; dx$

Step 6. Integrate both sides:

$\int \text{function of } u ; du = \int \text{function of } x ; dx$

Step 7. Substitute back $u = \frac{y}{x}$ (or $y = ux$) to get the final solution in terms of $x$ and $y$.

Step 8. If an initial condition is given (e.g. $y(x_0) = y_0$), use it to find the constant of integration.

Question 1:

Solve the homogeneous differential equation $\displaystyle \frac{dy}{dx} = \frac{x + y}{x – y},$.

Step 1. Let $u = \frac{y}{x} \Rightarrow y = ux$

Differentiate: $\displaystyle \frac{dy}{dx} = u + x\frac{du}{dx}$

Step 2. Substitute into the original equation:

$\displaystyle u + x\frac{du}{dx} = 1 + u$

Subtract $u$ from both sides: $\displaystyle x\frac{du}{dx} = 1$

Step 3. Separate and integrate:

$\displaystyle \frac{du}{dx} = \frac{1}{x} \Rightarrow \int du = \int \frac{1}{x},dx$

$\displaystyle u = \ln x + c = \ln(kx)$

Step 4. Substitute back $u = \frac{y}{x}$:

$\displaystyle \frac{y}{x} = \ln(kx) \Rightarrow y = x\ln(kx)$

or rearranged as: ${kx = e^{y/x}}$

Final Answer: ${kx = e^{,y/x}}$

This matches your handwritten solution exactly. ✅

Question 2:

Solve the homogeneous differential equation $\displaystyle \frac{dy}{dx} = \frac{y^2 – 2x^2}{xy}$,
where $x, y \ne 0$. Given that $y = 2$ when $x = 1$, find the solution in the form $y^2 = f(x)$.

Step 1. Use the substitution $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$

Substitute into the equation:

$\displaystyle v + x\frac{dv}{dx} = \frac{v^2x^2 – 2x^2}{x \cdot vx} = \frac{v^2 – 2}{v}$

Step 2. Rearranging:

$\displaystyle x\frac{dv}{dx} = \frac{v^2 – 2}{v} – v = \frac{v^2 – 2 – v^2}{v} = \frac{-2}{v}$

So: $\displaystyle \frac{dv}{dx} = \frac{-2}{vx}$

Step 3. Separate variables: $\displaystyle v,dv = -\frac{2}{x},dx$

Step 4. Integrate both sides: $\displaystyle \int v,dv = -2 \int \frac{1}{x},dx$

$\displaystyle \frac{v^2}{2} = -2\ln|x| + C$

Step 5. Back-substitute $v = \frac{y}{x}$:

$\displaystyle \frac{1}{2} \cdot \frac{y^2}{x^2} = -2\ln|x| + C$

Multiply both sides by $2x^2$:

$\displaystyle y^2 = 2x^2\left(-2\ln|x| + C\right)$

Step 6. Use the initial condition $y = 2$ when $x = 1$:

$\displaystyle 4 = 2(1)^2(-2\ln|1| + C) \Rightarrow 4 = 2(0 + C) \Rightarrow C = 2$

Final Answer: ${y^2 = 2x^2\left(-2\ln|x| + 2\right) = 4x^2(1 – \ln|x|)}$

Question 3:

Find the general solution of the differential equation $\displaystyle 2xy,\frac{dy}{dx} = 3y^2 – x^2$

Step 1. Rewrite the equation:

$\displaystyle 2,\frac{dy}{dx} = \frac{3y^2 – x^2}{x y}$

Express right-hand side in terms of $\frac{y}{x}$:

$\displaystyle 2,\frac{dy}{dx} = 3\left(\frac{y}{x}\right) – \frac{x}{y}$

Step 2. Use the substitution $u = \frac{y}{x} \Rightarrow y = ux$

Then: $\displaystyle \frac{dy}{dx} = u + x\frac{du}{dx}$

Substitute into the equation: $\displaystyle 2(u + x\frac{du}{dx}) = 3u – \frac{1}{u}$

Step 3. Solve for $\frac{du}{dx}$:

$\displaystyle 2x\frac{du}{dx} = 3u – \frac{1}{u} – 2u = \frac{u^2 – 1}{u}$

So: $\displaystyle \frac{du}{dx} = \frac{u^2 – 1}{2ux}$

Step 4. Separate variables: $\displaystyle \frac{2u}{u^2 – 1},du = \frac{dx}{x}$

Step 5. Integrate both sides: $\displaystyle \int \frac{2u}{u^2 – 1},du = \int \frac{dx}{x}$

Left side is: $\displaystyle \ln|u^2 – 1|$,

Right side is: $\ln|x| + C$

So: $\ln|u^2 – 1| = \ln|kx|$

Step 6. Remove logarithms: $u^2 – 1 = kx$

Substitute back $u = \frac{y}{x}$:

$\displaystyle \left(\frac{y}{x}\right)^2 – 1 = kx \Rightarrow \frac{y^2}{x^2} – 1 = kx$

Multiply through by $x^2$: $y^2 = kx^3 + x^2$

Final Answer: ${y^2 = kx^3 + x^2}$ ✅

Question 4:

Solve the differential equation $\displaystyle \frac{dy}{dx} = \frac{3y^2 + x^2}{2xy}$, given that $y(1) = 2$.

Find the particular solution in the form $y^2 + x^2 = f(x)$.

Step 1. Rearranging:

$\displaystyle \frac{dy}{dx} = \frac{3y}{2x} + \frac{x}{2y}$

Step 2. Let $u = \frac{y}{x} \Rightarrow y = ux$,

so $\displaystyle \frac{dy}{dx} = u + x\frac{du}{dx}$

Substitute into the equation:

$\displaystyle 2(u + x\frac{du}{dx}) = 3u + \frac{1}{u}$

Step 3. Simplify and solve for $\frac{du}{dx}$:

$\displaystyle 2x\frac{du}{dx} = \frac{u^2 + 1}{u}$

Step 4. Separate variables:

$\displaystyle \frac{2u}{u^2 + 1},du = \frac{dx}{x}$

Step 5. Integrate both sides:

$\displaystyle \int \frac{2u}{u^2 + 1},du = \int \frac{dx}{x}$

gives $\ln(u^2 + 1) = \ln x + C = \ln(kx)$

Step 6. Substitute back $u = \frac{y}{x}$:

$\displaystyle \ln\left(\frac{y^2}{x^2} + 1\right) = \ln(kx)
\Rightarrow \frac{y^2 + x^2}{x^2} = kx
\Rightarrow y^2 + x^2 = kx^3$

Step 7. Use the initial condition $y(1) = 2$:

$\displaystyle 4 + 1 = k(1)^3 \Rightarrow k = 5$

Final Answer (Particular Solution): ${y^2 + x^2 = 5x^3}$ ✅

Question 5:

Solve the differential equation $x,\frac{dy}{dx} = y + x,e^{y/x}, \quad \text{with } y(1) = 2$.

Solution:

Divide both sides by $x$:

$\displaystyle \frac{dy}{dx} = \frac{y}{x} + e^{y/x}$

Use the substitution $y = ux \Rightarrow \frac{dy}{dx} = u + x\frac{du}{dx}$

Now substitute into the equation:

$u + x\frac{du}{dx} = u + e^u$

Simplify:

$x\frac{du}{dx} = e^u$

Rewriting:

$e^{-u} \frac{du}{dx} = \frac{1}{x}$

Now integrate both sides:

$\displaystyle \int e^{-u} \frac{du}{dx} , dx = \int \frac{1}{x} , dx$

$\Rightarrow -e^{-u} = \ln x + c$

Substitute back $u = \frac{y}{x}$:

$-e^{-y/x} = \ln x + c$

Use the initial condition $y(1) = 2$:

$-e^{-2} = \ln(1) + c \Rightarrow -e^{-2} = 0 + c \Rightarrow c = -e^{-2}$

Therefore, the implicit solution is: $\ln x = e^{-2} – e^{-y/x}$

Question 6:

Use the substitution $u = x + y + 1$ to show that the differential equation $\displaystyle \frac{dy}{dx} = \frac{1}{x + y + 1}$

can be reduced to the differential equation $\displaystyle \frac{du}{dx} – 1 = \frac{1}{u}$.

Hence, show that the general solution is given by $x + y + 2 = ke^y$.

Step 1. Let $u = x + y + 1$, then

$\displaystyle \frac{du}{dx} = \frac{d}{dx}(x + y + 1) = 1 + \frac{dy}{dx}$

Step 2. Rearranging this, we get:

$\displaystyle \frac{dy}{dx} = \frac{du}{dx} – 1$

Step 3. The original differential equation is:

$\displaystyle \frac{dy}{dx} = \frac{1}{x + y + 1} = \frac{1}{u}$

Substitute:

$\displaystyle \frac{du}{dx} – 1 = \frac{1}{u}$

So,

$\displaystyle \frac{du}{dx} = \frac{1 + u}{u}$

Step 4. Separate the variables:

$\displaystyle \frac{u}{1 + u} \frac{du}{dx} = 1$

So,

$\displaystyle \int \frac{u}{1 + u} , \frac{du}{dx} , dx = \int dx$

Step 5. Simplify the integrand using algebra:

$\displaystyle \frac{u}{1 + u} = 1 – \frac{1}{1 + u}$

So the integral becomes:

$\displaystyle \int \left(1 – \frac{1}{1 + u}\right) du = \int dx$

Step 6. Integrate both sides:

$u – \ln|1 + u| = x + c$

Step 7. Solve for $u$:

$\ln|1 + u| = u – x – c$

Then,

$1 + u = e^{u – x – c} = ke^{u – x}$ (where $k = e^{-c}$)

Step 8. Substitute back $u = x + y + 1$:

$1 + x + y + 1 = ke^{y + 1}$

$x + y + 2 = ke^y$

Final General Solution: $\displaystyle x + y + 2 = ke^y$

Question 7:

Use the substitution $y = ux$ to show that the differential equation

$\displaystyle x \frac{dy}{dx} = y^2 + x^2 + y$

can be reduced to the differential equation. $\displaystyle \frac{du}{dx} = u^2 + 1$.

Hence, show that the general solution is $\displaystyle y = x \tan(x + c)$.

Solution:

Step 1. Let $y = ux$, so

$\displaystyle \frac{dy}{dx} = u + x \frac{du}{dx}$

Step 2. Substitute into the given equation:

$\displaystyle x \left( u + x \frac{du}{dx} \right) = (ux)^2 + x^2 + ux$

Step 3. Expand both sides:

Left-hand side: $\displaystyle xu + x^2 \frac{du}{dx}$

Right-hand side: $\displaystyle u^2x^2 + x^2 + ux$

Step 4. Subtract $xu$ from both sides:

$\displaystyle x^2 \frac{du}{dx} = u^2x^2 + x^2 + ux – xu$

$\Rightarrow x^2 \frac{du}{dx} = u^2x^2 + x^2$

Step 5. Divide both sides by $x^2$:

$\displaystyle \frac{du}{dx} = u^2 + 1$

Step 6. Solve the differential equation by separating variables:

$\displaystyle \frac{du}{u^2 + 1} = dx$

Step 7. Integrate both sides:

$\displaystyle \int \frac{1}{u^2 + 1} , du = \int dx$

$\Rightarrow \arctan(u) = x + c$

Step 8. Substitute back $u = \frac{y}{x}$:

$\displaystyle \arctan\left( \frac{y}{x} \right) = x + c$

Step 9. Take tangent of both sides:

$\displaystyle \frac{y}{x} = \tan(x + c)$

$\Rightarrow y = x \tan(x + c)$

Final General Solution: $\displaystyle y = x \tan(x + c)$

Question 8:

Solve the homogeneous differential equation

$\displaystyle x \frac{dy}{dx} = y + x^2 \sin x$, given that $\left( \frac{\pi}{2}, \frac{\pi}{2} \right)$ lies on the curve.

Step 1. Let $y = ux$, so

$\displaystyle \frac{dy}{dx} = u + x \frac{du}{dx}$

Step 2. Substitute into the differential equation:

$\displaystyle x \left( u + x \frac{du}{dx} \right) = ux + x^2 \sin x$

Step 3. Expand both sides:

Left-hand side:

$\displaystyle xu + x^2 \frac{du}{dx}$

Right-hand side:

$\displaystyle ux + x^2 \sin x$

Step 4. Subtract $xu$ from both sides:

$\displaystyle x^2 \frac{du}{dx} = x^2 \sin x$

Step 5. Divide both sides by $x^2$:

$\displaystyle \frac{du}{dx} = \sin x$

Step 6. Integrate both sides:

$\displaystyle \int du = \int \sin x , dx$

$\Rightarrow u = -\cos x + c$

Step 7. Substitute back $u = \frac{y}{x}$:

$\displaystyle \frac{y}{x} = -\cos x + c$

$\Rightarrow y = x(-\cos x + c)$

Step 8. Use the initial condition $\left( \frac{\pi}{2}, \frac{\pi}{2} \right)$:

$\displaystyle \frac{\pi}{2} = \frac{\pi}{2}(-\cos(\frac{\pi}{2}) + c)$

$\Rightarrow \frac{\pi}{2} = \frac{\pi}{2}(0 + c) \Rightarrow c = 1$

Final Particular Solution: $\displaystyle y = x(1 – \cos x)$

Question 9:

Solve the differential equation $\displaystyle \frac{dy}{dx} = \frac{y^2 + 3xy + 2x^2}{x^2}$ given that $y = -1$ when $x = 1$.

Give your answer in the form $y = f(x)$.

Step 1. Use the substitution $y = vx$, so that

$\displaystyle \frac{dy}{dx} = v + x \frac{dv}{dx}$

Step 2. Substitute into the equation:

$\displaystyle v + x \frac{dv}{dx} = \frac{(vx)^2 + 3x(vx) + 2x^2}{x^2}$

Step 3. Simplify the right-hand side:

$\displaystyle \frac{v^2 x^2 + 3vx^2 + 2x^2}{x^2} = v^2 + 3v + 2$

Step 4. Substitute back:

$\displaystyle v + x \frac{dv}{dx} = v^2 + 3v + 2$

Step 5. Subtract $v$ from both sides:

$\displaystyle x \frac{dv}{dx} = v^2 + 2v + 2$

Step 6. Separate the variables:

$\displaystyle \int \frac{dv}{v^2 + 2v + 2} = \int \frac{dx}{x}$

Step 7. Complete the square in the denominator:

$v^2 + 2v + 2 = (v + 1)^2 + 1$

$\Rightarrow \int \frac{dv}{(v + 1)^2 + 1} = \ln|x| + c$

Step 8. Use the standard integral result:

$\Rightarrow \arctan(v + 1) = \ln x + c$

Step 9. Substitute back $v = \frac{y}{x}$:

$\displaystyle \arctan\left( \frac{y}{x} + 1 \right) = \ln x + c$

Step 10. Use the initial condition $x = 1$, $y = -1$:

$\displaystyle \arctan(0 + 1) = \ln(1) + c$

$\Rightarrow \arctan(1) = 0 + c \Rightarrow c = \frac{\pi}{4}$

Step 11. Final implicit equation:

$\displaystyle \arctan\left( \frac{y}{x} + 1 \right) = \ln x + \frac{\pi}{4}$

Step 12. Solve for $y$:

$\displaystyle \frac{y}{x} + 1 = \tan\left( \ln x + \frac{\pi}{4} \right)$

$\Rightarrow y = x \left( \tan\left( \ln x + \frac{\pi}{4} \right) – 1 \right)$

Final Answer: $\displaystyle y = x \left( \tan\left( \ln x + \frac{\pi}{4} \right) – 1 \right)$

Question 10:

Solve the differential equation $\displaystyle \frac{dy}{dx} = y^2 + xy + 4x^2$, given that $y = 2$ when $x = 1$.

Give your answer in the form $y = f(x)$.

Solution:

Step 1. Use the substitution $y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$

Step 2. Substitute into the differential equation:

$\displaystyle v + x \frac{dv}{dx} = v^2 x^2 + vx^2 + 4x^2 = x^2 (v^2 + v + 4)$

Step 3. Divide both sides by $x$:

$\displaystyle \frac{dv}{dx} = x (v^2 + v + 4)$

Step 4. Separate variables:

$\displaystyle \int \frac{dv}{v^2 + v + 4} = \int x , dx$

Step 5. Complete the square in the denominator:

$v^2 + v + 4 = \left(v + \frac{1}{2}\right)^2 + \frac{15}{4}$

So we integrate:

$\displaystyle \int \frac{dv}{\left(v + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{15}}{2}\right)^2} = \int x , dx$

This gives:

$\displaystyle \frac{2}{\sqrt{15}} \tan^{-1} \left( \frac{2v + 1}{\sqrt{15}} \right) = \frac{x^2}{2} + C$

Step 6. Use the initial condition $y = 2$, $x = 1$

Then $v = \frac{y}{x} = 2$

$\displaystyle \frac{2}{\sqrt{15}} \tan^{-1} \left( \frac{2(2) + 1}{\sqrt{15}} \right) = \frac{1}{2} + C$

$\Rightarrow C = \frac{2}{\sqrt{15}} \tan^{-1} \left( \frac{5}{\sqrt{15}} \right) – \frac{1}{2}$

Step 7. Final answer in the form $y = f(x)$:

$\displaystyle \frac{2}{\sqrt{15}} \tan^{-1} \left( \frac{2y/x + 1}{\sqrt{15}} \right) = \frac{x^2}{2} + C$

or in simplified form:

$\displaystyle y = x \left( \frac{\sqrt{15}}{2} \tan\left( \frac{x^2}{2} + C \right) – \frac{1}{2} \right)$

General Steps to Solve a Homogeneous Differential Equation

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