Question

The general solution of the differential equation x²dy - 2xydx = x⁴cosx dx is

• A. y = x²sin x + cx²
• B. y = x²sin x + c
• C. y = sin x + cx²
• D. y = cos x + cx²

Answer 1

The Correct Option is A

We are given the differential equation \(x^2 dy - 2xy dx = x^4 \cos x \, dx\) and we need to find the general solution.

First, rewrite the equation as:

\(x^2 dy = 2xy dx + x^4 \cos x \, dx\)

Divide both sides by \(x^2\):

\(dy = \frac{2xy}{x^2} dx + \frac{x^4}{x^2} \cos x \, dx\)

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

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

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

This is a first-order linear differential equation in the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = -\frac{2}{x}\) and \(Q(x) = x^2 \cos x\).

Find the integrating factor \(I(x) = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = e^{\ln x^{-2}} = x^{-2} = \frac{1}{x^2}\).

Multiply the differential equation by the integrating factor:

\(\frac{1}{x^2} \frac{dy}{dx} - \frac{2}{x^3} y = \cos x\)

\(\frac{d}{dx} (\frac{1}{x^2} y) = \cos x\)

Integrate both sides with respect to x:

\(\int \frac{d}{dx} (\frac{1}{x^2} y) dx = \int \cos x \, dx\)

\(\frac{y}{x^2} = \sin x + c\)

\(y = x^2 \sin x + c x^2\)

Therefore, the general solution of the differential equation is \(y = x^2 \sin x + c x^2\).

Thus, the correct option is (A) \(y = x^2 \sin x + cx^2\).

Answer 2

Rewriting the equation:

$$ x^2 dy - 2xy dx = x^4 \cos x dx \implies dy - \frac{2}{x} y dx = x^2 \cos x dx. $$

This is a first-order linear differential equation. The integrating factor is:

$$ e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = e^{\ln x^{-2}} = x^{-2} = \frac{1}{x^2}. $$

Multiply through by $ \frac{1}{x^2} $:

$$ \frac{1}{x^2} dy - \frac{2}{x^3} y dx = \cos x dx \implies d\left(\frac{y}{x^2}\right) = \cos x dx. $$

Integrate both sides:

$$ \frac{y}{x^2} = \int \cos x dx = \sin x + c \implies y = x^2 \sin x + cx^2. $$

Final Answer: The final answer is $ {y = x^2 \sin x + cx^2} $.