Question

The particular solution of the differential equation, \[ x y \frac{dy}{dx} = x^2 + 2y^2 \quad \text{when} \quad y(1) = 0 \quad \text{is:} \]

• A. \( \frac{x^2 + y^2}{x^3} = 1 \)
• B. \( x^2 + y^2 = x \)
• C. \( x^2 + 2y^2 = x^4 \)
• D. \( x^2 + y^2 = x^4 \)

Hint:

For solving differential equations, always ensure you separate variables, integrate each side, and apply the initial condition carefully to get the particular solution.

Answer 1

The Correct Option is D

We are given: 

\( x y \frac{dy}{dx} = x^2 + 2y^2 \)

Divide both sides by \( x y \):

\( \frac{dy}{dx} = \frac{x^2 + 2y^2}{x y} \)

Split the RHS:

\( \frac{dy}{dx} = \frac{x}{y} + \frac{2y}{x} \)

Now multiply both sides by \( y \) to simplify:

\( y \frac{dy}{dx} = x + \frac{2y^2}{x} \)

This is still messy, so let’s try substitution. Let’s try:

Let \( z = y^2 \Rightarrow \frac{dz}{dx} = 2y \frac{dy}{dx} \)

From the original equation:

\( x y \frac{dy}{dx} = x^2 + 2y^2 \Rightarrow x \cdot \frac{1}{2} \cdot \frac{dz}{dx} = x^2 + 2z \)

So:

\( \frac{1}{2} x \frac{dz}{dx} = x^2 + 2z \)

Multiply both sides by 2:

\( x \frac{dz}{dx} = 2x^2 + 4z \)

Rewriting:

\( \frac{dz}{dx} - \frac{4z}{x} = 2x \)

This is a linear differential equation in \( z \). Use integrating factor (IF):

\( \text{IF} = e^{\int -\frac{4}{x} dx} = x^{-4} \)

Multiplying both sides by IF:

\( x^{-4} \frac{dz}{dx} - \frac{4z}{x^5} = 2x^{-3} \)

This simplifies to:

\( \frac{d}{dx}(z x^{-4}) = 2x^{-3} \)

Integrate both sides:

\( z x^{-4} = \int 2x^{-3} dx = -x^{-2} + C \)

So:

\( z = x^4 (-x^{-2} + C) = -x^2 + Cx^4 \)

Recall \( z = y^2 \), so:

\( y^2 = -x^2 + Cx^4 \Rightarrow x^2 + y^2 = Cx^4 \)

Now use the initial condition \( y(1) = 0 \):

\( x = 1, y = 0 \Rightarrow 1^2 + 0^2 = C(1)^4 \Rightarrow C = 1 \)

Final solution:

\( \boxed{x^2 + y^2 = x^4} \)