Question
The general solution of the differential equation $$\frac{{dy}}{{dx}} - \frac{{\tan \,y}}{{1 + x}} = \left( {1 + x} \right){e^x}\sec \,y$$ is :
A.
$$\sin \left( {1 + x} \right) = y\left( {{e^x} + c} \right)$$
B.
$$y\,\sin \left( {1 + x} \right) = c{e^x}$$
C.
$$\left( {1 + x} \right)\sin \,y = {e^x} + c$$
D.
$$\sin \,y = \left( {1 + x} \right)\left( {{e^x} + c} \right)$$
Answer :
$$\sin \,y = \left( {1 + x} \right)\left( {{e^x} + c} \right)$$
Solution :
Divide the equation by $$\sec \,y$$
$$\eqalign{
& \cos \,y\frac{{dy}}{{dx}} - \frac{{\sin \,y}}{{1 + x}} = \left( {1 + x} \right){e^x} \cr
& {\text{Put }}\sin \,y = z \Rightarrow \cos \,y\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}{\text{ then}} \cr
& \frac{{dz}}{{dx}} - \left( {\frac{1}{{1 + x}}} \right)z = \left( {1 + x} \right){e^x} \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{ - \int {\frac{1}{{1 + x}}dx} }} = {e^{ - \log \left( {1 + x} \right)}} = \frac{1}{{1 + x}} \cr} $$
The solution is $$z\left( {\frac{1}{{1 + x}}} \right) = {e^x} + c \Rightarrow \sin \,y = \left( {1 + x} \right)\left( {{e^x} + c} \right)$$