Find the general solution for each of the following differential equations. $\frac{d y}{d x}-y \tan x=e^{x} \sec x$

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CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal

Find the general solution for each of the following differential equations. $\frac{d y}{d x}-y \tan x=e^{x} \sec x$

Solution:

$\frac{d y}{d x}-y \tan x=e^{x} \sec x-\ldots(1)$
To solve (1) we will use following formula
$\begin{array}{l} \quad \int \tan x d x=\log (\sec x) \\ a \log b=\log b^{a} \\ a^{\log _{a} b}=b \\ \int e^{x} d x=e^{x} \end{array}$
General solution for the differential equation in the form of
$\frac{d y}{d x}+P y=Q$
General solution is given by,
$y \cdot\left(I . F_{.}\right)=\int Q \cdot\left(I . F_{\cdot}\right) d x+c$
Where, integrating factor,
$\text { I. } F .=e \int^{p d x}$
Equation (1) is of the form, Comparing it with
$\frac{d y}{d x}+P y=Q$
Where, $P=-\tan x$ and $Q=e^{x} \sec x$
Therefore, integrating factor is
$\begin{array}{c} \text { I.F. }=e \int^{p d x} \\ e^{-\tan x d x} \\ e^{-\log (\sec x)} \ldots \ldots \ldots . .\left(\int \tan x d x=\log (\sec x)\right. \end{array}$
$\begin{array}{l} e^{\log (\sec x)^{-1}} \ldots \ldots \ldots \ldots\left(a \log b=\log b^{a}\right) \\ e^{\log (\cos x)} \\ \cos x \ldots \ldots \ldots \ldots\left(a^{\log _{a} b}=b\right) \end{array}$
General solution is
$\begin{array}{r} y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c \\ y \cdot(\cos x)=\int\left(e^{x} \sec x\right) \cdot(\cos x) d x+c \\ y \cdot(\cos x)=\int\left(e^{x} \frac{1}{\cos x}\right) \cdot(\cos x) d x+c \\ y \cdot(\cos x)=\int e^{x} d x+c \\ \text { y. }(\cos x)=e^{x}+c \ldots \ldots \ldots \ldots \ldots\left(\int e^{x} d x=e^{x}\right. \end{array}$
So, general solution is
$\mathrm{y} \cdot(\cos \mathrm{x})=e^{x}+c$