Question

The solution of the differential equation $(x + 1) \frac{dy}{dx} - y = e^{3x} (x + 1)^2$ is

• A. y = (x + 1)e^{3x} + C
• B. 3y = (x + 1) + e^{3x} + C
• C. \frac{3y}{x+1} = e^{3x} + C
• D. ye^{-3x} = 3(x + 1) + C

Answer 1

Answer: (C)

\frac{3y}{x+1} = e^{3x} + C

Answer 2

The given differential equation is $(x + 1) \frac{dy}{dx} - y = e^{3x} (x + 1)^2$. This is a first-order linear differential equation. We rewrite it in the standard form: $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide the entire equation by $(x+1)$: $\frac{dy}{dx} - \frac{1}{x+1} y = e^{3x} (x + 1)$

Comparing with the standard form: $P(x) = -\frac{1}{x+1}$ $Q(x) = e^{3x} (x + 1)$

The integrating factor (IF) is $IF = e^{\int P(x) dx}$. $IF = e^{\int -\frac{1}{x+1} dx} = e^{-\ln|x+1|} = \frac{1}{x+1}$.

The general solution is $y \times IF = \int Q(x) \times IF \, dx + C$. $y \times \frac{1}{x+1} = \int e^{3x} (x + 1) \times \frac{1}{x+1} \, dx + C$ $\frac{y}{x+1} = \int e^{3x} \, dx + C$ $\int e^{3x} \, dx = \frac{e^{3x}}{3}$

So, the solution is: $\frac{y}{x+1} = \frac{e^{3x}}{3} + C_{initial}$ Multiply both sides by 3: $\frac{3y}{x+1} = e^{3x} + 3C_{initial}$ Let $C = 3C_{initial}$. Thus, the general solution is: $\frac{3y}{x+1} = e^{3x} + C$