Question

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

Hint:

For a linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x), \] first compute the integrating factor \[ \text{I.F.=e^{\int P(x)\,dx}. \] Multiplying the equation by the I.F. converts the left side into \(\frac{d}{dx}(y\cdot \text{I.F.)\), making the equation easy to integrate.

Answer 1

Concept: The given equation is a linear differential equation of the form \[ \frac{dy}{dx}+P(x)y=Q(x) \] where \[ P(x)=\frac{1}{x}, \qquad Q(x)=x^2 \] The solution is obtained using the integrating factor (I.F.): \[ \text{I.F.=e^{\int P(x)\,dx} \]
Step 1: Find the integrating factor. \[ \text{I.F.=e^{\int \frac{1}{x}dx} \] \[ =e^{\ln x}=x \]
Step 2: Multiply the equation by the integrating factor. \[ x\frac{dy}{dx}+y=x^3 \] The left-hand side becomes the derivative of a product: \[ \frac{d}{dx}(xy)=x^3 \]
Step 3: Integrate both sides. \[ xy=\int x^3\,dx \] \[ xy=\frac{x^4}{4}+C \]
Step 4: Express the solution for \(y\). \[ y=\frac{x^4}{4x}+\frac{C}{x} \] \[ y=\frac{x^3}{4}+\frac{C}{x} \]
Step 5: General solution. \[ \boxed{y=\frac{x^3}{4}+\frac{C}{x}} \]