Question

The solution of the differential equation $\frac{dy}{dx}-y\tan \left(x\right)=e^x\sec \left(x\right)$ is

• A. $y=e^x\cos \left(x\right)+c$
• B. $y\cos \left(x\right)=e^x+c$
• C. $y=e^x\sin \left(x\right)+c$
• D. $y\sin \left(x\right)=e^x+c$

Answer 1

The correct option is B

$y\cos \left(x\right)=e^x+c$

Explanation of the correct option.

Compute the required value.

Given : $\frac{dy}{dx}-y\tan \left(x\right)=e^x\sec \left(x\right)$

Compare the equation with standard L.D.E.$\frac{dy}{dx}+Py=Q$,

$P=-\tan \left(x\right)$ and $Q=e^x\sec \left(x\right)$

$\begin{array}{rcl}\mathrm{I}.\mathrm{F}.& =& {\mathrm{e}}^{\int \mathrm{P}\mathrm{dx}}\\ & =& {\mathrm{e}}^{\int -\tan \left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{-\log \left(\left|\sec \left(\mathrm{x}\right)\right|\right)}\\ & =& {\mathrm{e}}^{\log \left(\cos \left(\mathrm{x}\right)\right)}\\ & =& \cos \left(\mathrm{x}\right)\end{array}$

Since the solution of L.D.E. is given by $y(\mathrm{I}.\mathrm{F}.)=\int Q\left(\mathrm{I}.\mathrm{F}.\right)dx+c$.

$$\begin{gathered} y\cos \left(x\right)=\int e^x\sec \left(x\right)\cos \left(x\right)dx+c \\ \Rightarrow y\cos \left(x\right)=\int e^{x}dx+c \\ \Rightarrow y\cos \left(x\right)=e^x+c \end{gathered}$$

Hence option $\left(B\right)$ is the correct option.