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Solve the following differential equation : $\frac{dx}{dy}+x= \left ( \tan y+\sec ^{2}y \right )$

Answers (1)

R Ravindra Pindel

The given D.E is of the form $\frac{dx}{dy}+p\left ( y \right )x= Q\left ( y \right )$
where $p\left ( y \right )= 1\; \S \; Q\left ( y \right )= \tan y+\sec ^{2}y$
Now IF $=e^ {\int 1dy}= e^{y}$
The solution is given as $xe^{y}= \int \left ( \tan y+\sec ^{2}y \right )dy$
ie $xe^{y}= e^{y}\left ( \tan y \right )+c\; or\; x= \tan y+e^{-y}c.$

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