Solve the following differential equation: dy dx +y= - Vidyadip

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}=\cos\text{x}$

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Answer

Here, $\frac{\text{dy}}{\text{dx}}+\text{y}=\cos\text{x}$
It is a linear differential equation. Comparing the equation by,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=1,\text{Q}=\cos\text{x}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\text{dx}}$
$=\text{e}^{\text{x}}$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}(\text{e}^{\text{x}})=\int(\cos\text{x})(\text{e}^{\text{x}})+\text{C}_1\ \dots(\text{i})$
Let $\text{I}=\int\text{e}^{\text{x}}\cos\text{xdx}$
$=\cos\text{x}\times\int\text{e}^{\text{x}}\text{dx}\int(\sin\text{x}\int\text{e}^{\text{x}}\text{dx})\text{dx + C}_2$
Using integration by parts
$\text{I}=\text{e}^{\text{x}}\cos{\text{x}}+\int\sin\text{xe}^{\text{x}}\text{dx + C}$
$=\text{e}^{\text{x}}\cos\text{x}+\big[\sin\text{x}\int\text{e}^{\text{x}}\text{dx}-\int(\cos\text{x}\int\text{e}^{\text{x}}\text{dx})\text{dx}\big]+\text{C}_2$
$\text{I}=\text{e}^{\text{x}}\cos\text{x}+\sin\text{e}^\text{x}-\text{I}+\text{C}_2$
$2\text{I}=\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})+\text{C}_2$
$\text{I}=\frac{\text{e}^{\text{x}}}{2}(\cos\text{x}+\sin\text{x})+\frac{\text{C}_2}{2}$
$\text{I}=\frac{\text{e}^{\text{x}}}{2}(\cos\text{x}+\sin\text{x})+\text{C}_3$
Putting I in equation (i),
$\text{ye}^{\text{x}}=\frac{\text{e}^{\text{x}}}{2}(\cos\text{x}+\sin{\text{x}})+\text{C}_1+\text{C}_3$
$\text{ye}^{\text{x}}=\frac{\text{e}^{\text{x}}}2(\cos\text{x}+\sin\text{x})+\text{C}$
$\text{y}=\frac{1}2(\cos\text{x}+\sin\text{x})+\text{Ce}^{-\text{x}}$

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