Question

Solve the following differential equation :-


$\longrightarrow \bf{ \cos {}^{2} x \frac{DY}{DX} + y = \tan \: x}$
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Answer 1

$\huge\mathfrak{\bf{\underline{\underline{Question \ :}}}}$

$\longrightarrow\bf{ \cos {}^{2} x \frac{DY}{DX} + y = \tan \: x}$

$\huge\mathfrak{\bf{\underline{\underline{\blue{Answer \ :}}}}}$

$\implies \: \mathtt{ \frac{dy}{dx} + (sec^{2} x)y = tanx. {sec}^{2} x}$

$\implies \: \mathtt{P = {sec}^{2}x \: \: \: \: \: \: \: \: \: Q = {sec}^{2}x.tanx }$

$\implies \: \mathtt{If = \: \: {e}^{ \int \: (P \: dx)} } = \: \: \mathtt{ {e}^{ \int \:( {sec}^{2}x \: dx )} }$

$\implies \: \mathtt{ Y(If) = \int{Q(If)dx + C}}$

$\implies \: \mathtt{Y. {e}^{tanx} = \int{tanx . {sec}^{2}x. {e}^{tanx}dx }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\underline{Put \: \: \: \mathtt{t = tanx}}}$

$\implies \: \mathtt{Y. {e}^{tanx} = \mathtt{ \int{t. {e}^{t} dt + C}}} \: \: \: ..... \mathsf{LIATE \: Rule}$

$\implies \: \mathtt{Y. {e}^{tanx} = t. {e}^{t} - 1( {e}^{t} ) + C}$

$\implies \: \mathtt{Y. {e}^{tanx} = tanx. {e}^{tanx} - {e}^{tanx} + C }$

$\mathfrak{Answer}:- \: \boxed{ \mathtt \blue{Y. {e}^{tanx} = {e}^{tanx} (tanx - 1) +C }}$

Answer 2

Step-by-step explanation:

$i \: hope \: itz \: help \: \: uh$