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The solution of the differential equation
$(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}$ is
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$(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}$ is
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Verified Answer
The correct answer is:
$\frac{3 y}{x+1}=e^{3 x}+c$
The given equation is $\frac{d y}{d x}-\frac{y}{x+1}=e^{3 x}(x+1)$
I.F. $=e^{\int-\frac{1}{x+1} d x}=e^{-\log (x+1)}=\frac{1}{x+1}$
The solution is
$y\left(\frac{1}{x+1}\right)=\int e^{3 x}(x+1) \cdot \frac{1}{x+1} d x+a$
$\Rightarrow \frac{y}{x+1}=\int e^{3 x} d x+a=\frac{e^{3 x}}{3}+a$
$\Rightarrow \frac{3 y}{x+1}=e^{3 x}+c, c=3 a$
I.F. $=e^{\int-\frac{1}{x+1} d x}=e^{-\log (x+1)}=\frac{1}{x+1}$
The solution is
$y\left(\frac{1}{x+1}\right)=\int e^{3 x}(x+1) \cdot \frac{1}{x+1} d x+a$
$\Rightarrow \frac{y}{x+1}=\int e^{3 x} d x+a=\frac{e^{3 x}}{3}+a$
$\Rightarrow \frac{3 y}{x+1}=e^{3 x}+c, c=3 a$
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