Question

The general solution of the differential equation \[ \frac{dy}{dx}+\frac{y}{x}=x^2 \] is

• A. \[ y=\frac13x^3+\frac{C}{x} \]
• B. \[ y=\frac14x^4+Cx \]
• C. \[ y=\frac14x^3+C \]
• D. \[ y=\frac14x^3+\frac{C}{x} \]

Hint:

For a linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x), \] use \[ \text{I.F.}=e^{\int P(x)\,dx}. \] Then \[ y\cdot \text{I.F.} = \int Q(x)\,\text{I.F.}\,dx + C. \]

Answer 1

The Correct Option is D

Concept: The equation is a first-order linear differential equation of the form \[ \frac{dy}{dx}+P(x)y=Q(x). \] Here, \[ P(x)=\frac1x, \qquad Q(x)=x^2. \]

Step 1: Find the Integrating Factor (I.F.). \[ \text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac1x\,dx} = e^{\ln x} = x. \]

Step 2: Multiply the equation by the I.F. \[ x\frac{dy}{dx}+y=x^3. \] The left side becomes \[ \frac{d}{dx}(xy). \] Hence \[ \frac{d}{dx}(xy)=x^3. \]

Step 3: Integrate both sides. \[ xy = \int x^3\,dx. \] \[ xy = \frac{x^4}{4}+C. \]

Step 4: Solve for \(y\). \[ y = \frac{x^4}{4x} + \frac{C}{x}. \] \[ y = \frac14x^3+\frac{C}{x}. \]

Step 5: Write the final answer. \[ \boxed{ y=\frac14x^3+\frac{C}{x} } \]