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BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Mathematics
Chapter: Differential Equations

BITSAT 2024 · Mathematics · Differential Equations

The solution of the differential equation (x+1) (d y)/(d x)-y=e^ 3...

The solution of the differential equation

(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^2

is

Options

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

Correct answer

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

Step-by-step solution

Given,

\frac{d y}{d x}-\frac{y}{x+1}=e^{3 x}(x+1)

Above is linear differential equation of form

\begin{aligned} & \frac{d y}{d x}+P x=Q \\ & \therefore \quad \mathrm{IF}=e^{\int P d x}=e^{\int-\frac{1}{x+1} d x}=e^{-\ln (1+x)}=\frac{1}{1+x} \end{aligned}

\Rightarrow

Solution will be

\begin{aligned} y \cdot \mathrm{IF} & =\int Q \cdot \mathrm{IF} d x \\ y \cdot \frac{1}{1+x} & =\int e^{3 x}(x+1) \times \frac{1}{(1+x)} d x \\ \frac{y}{1+x} & =\frac{e^{3 x}}{3}+C^{\prime} \\ \frac{3 y}{1+x} & =e^{3 x}+C^{\prime} \quad \quad\left[\because C=3 C^{\prime}\right] \end{aligned}

$

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