Find the general solution for each of the following differential equations. $\frac{d y}{d x}+y=\cos x-\sin x$

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

Find the general solution for each of the following differential equations. $\frac{d y}{d x}+y=\cos x-\sin x$

Solution:

$\frac{d y}{d x}+y=\cos x-\sin x-\cdots \dots(1)$
General solution for the differential equation in the form of
$\frac{d y}{d x}+P y=Q$
is given by,
$y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$
Where, integrating factor,
$\begin{array}{r} \text { I. } F .=e \int^{p d x} \\ \frac{d y}{d x}+\mathrm{y}=\cos \mathrm{x}-\sin \mathrm{x} \ldots \ldots \ldots(1) \end{array}$
Equation (1) is of the form
$\frac{d y}{d x}+P y=Q$
Where, $\mathrm{P}=1$ and $\mathrm{Q}=\operatorname{Cos} \mathrm{x}-\operatorname{Sin} \mathrm{x}$
Therefore, integrating factor is
$\text { I.F. }=e \int^{p d x}=e^{x}$
General solution is
$\begin{array}{r} y \cdot(I . F .)=\int Q \cdot(I . F \cdot) d x+c \\ \therefore \mathrm{y} \cdot\left(e^{x}\right)=\int(\cos x-\sin x) \cdot\left(e^{x}\right) d x+c \end{array}$
Let, $f(x)=\cos x=>f^{\prime}(x)=-\sin x$
$\therefore \mathrm{y} \cdot\left(e^{x}\right)=\left(e^{2 x}\right) \cdot \cos x+c$
Dividing above equation by $\mathrm{e}^{\mathrm{x}}$ ,
$\therefore y=\cos x+\frac{c}{e^{x}}$