Question
The solution to of the differential equation $$\left( {x + 1} \right)\frac{{dy}}{{dx}} - y = {e^{3x}}{\left( {x + 1} \right)^2}{\text{ is :}}$$
A.
$$y = \left( {x + 1} \right){e^{3x}} + c$$
B.
$$3y = \left( {x + 1} \right) + {e^{3x}} + c$$
C.
$$\frac{{3y}}{{x + 1}} = {e^{3x}} + c$$
D.
$$y{e^{ - 3x}} = 3\left( {x + 1} \right) + c$$
Answer :
$$\frac{{3y}}{{x + 1}} = {e^{3x}} + c$$
Solution :
The given equation is
$$\eqalign{
& \frac{{dy}}{{dx}} - \frac{y}{{x + 1}} = {e^{3x}}\left( {x + 1} \right) \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int { - \frac{1}{{x + 1}}dx} }} = {e^{ - \log \left( {x + 1} \right)}} = \frac{1}{{x + 1}} \cr
& {\text{The solution is }} \cr
& y\left( {\frac{1}{{x + 1}}} \right) = \int {{e^{3x}}\left( {x + 1} \right).\frac{1}{{x + 1}}dx + a} \cr
& \Rightarrow \frac{y}{{x + 1}} = \int {{e^{3x}}dx + a} \cr
& \Rightarrow \frac{y}{{x + 1}} = \frac{{{e^{3x}}}}{3} + a \cr
& \Rightarrow \frac{{3y}}{{x + 1}} = {e^{3x}} + c,\,c = 3a \cr} $$