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Find the general solution for each of the following differential equations. \frac{d y}{d x}=y \cot x=\sin 2 x
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 \cot x=\sin 2 x

Solution:

\frac{d y}{d x}+y \cot x=\sin 2 x-\dots (1)
General solution:
For the differential equation in the form of \frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}
General solution is given by,
y \cdot\left(I . F_{\cdot}\right)=\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}+y \cot x=\sin 2 x \ldots \ldots \ldots(1) \end{array}
Equation (1) is of the form
\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}
Where, P=\operatorname{Cot} x and Q=\sin 2 x
Therefore, integrating factor is
\text { I. } F .=e \int^{p d x}=\operatorname{Sin} \mathrm{x}
General solution is
\begin{array}{l} y \cdot(I . F .)=\int Q \cdot(I . F \cdot) d x+c \\ \therefore \mathrm{y} \cdot(\sin x)=\int(\sin 2 x) \cdot(\sin x) \mathrm{dx}+\mathrm{c} \ldots \ldots \ldots \ldots \ldots \\ \mathrm{I}=\int(\sin 2 x) \cdot(\sin x)^{\mathrm{dx}} \end{array}
Let, u=\sin 2 x \& v=\sin x
\therefore \mathrm{I}=\sin 2 x \cdot \int \sin x \mathrm{~d} \mathrm{x}-\int\left(\frac{d}{d t}(\sin 2 x) \cdot \int \sin x \mathrm{dx}\right) \mathrm{d} \mathrm{x}
\therefore I=-\sin 2 x \cdot \cos x-\int((2 \cos 2 x) \cdot(-\cos x))^{d x}
\therefore I=-\sin 2 x \cdot \cos x+2 \int((\cos 2 x) \cdot(\cos x))^{d x}
Again let, u=\cos 2 x \& v=\cos x
+2\left\{\cos 2 x \cdot \int \cos x d x-\int\left(\frac{d}{d t}(\cos 2 x) \cdot \int \cos x d x\right) d x\right\}
\therefore \mathrm{I}=-\sin 2 x \cdot \cos x+2\left\{\cos 2 x \cdot \sin x-\int((-2 \sin 2 x) \cdot(\sin x)) d x\right\}
\therefore \mathrm{I}=-\sin 2 x \cdot \cos x+2\left\{\cos 2 x \cdot \sin x+2 \int((\sin 2 x) \cdot(\sin x)) d x\right\}
\therefore I=-\sin 2 x \cdot \cos x+2\{\cos 2 x \cdot \sin x+2 I\}
\therefore I=-\sin 2 x \cdot \cos x+2 \cos 2 x \cdot \sin x+4 I
\therefore \mathrm{I}-4 \mathrm{I}=-2 \sin x \cos x \cdot \cos x+2\left(\cos ^{2} x-\sin ^{2} x\right) \cdot \sin x
\cdots \cdots \ldots\left(\because \sin 2 x=2 \sin x \cdot \cos x \& \cos 2 x=\left(\cos ^{2} x-\sin ^{2} x\right)\right)
\therefore-3 \mathrm{I}=-2 \sin x \cos ^{2} x+2 \sin x \cos ^{2} x-2 \sin ^{3} x
\therefore-3 I=-2 \sin ^{3} x
\therefore I=\frac{2}{3} \sin ^{3} x
Substituting I in (2),
\therefore \mathrm{y} \cdot(\sin x)=\frac{2}{3} \sin ^{3} x+\mathrm{c}
Therefore, general solution is
\text { y. }(\sin x)=\frac{2}{3} \sin ^{3} x+c

i

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