class 12 maths differential equations

Find the general solution of $\frac{{dy}}{{dx}} - 3y = \sin 2x$.

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📘 Differential Equations NCERT EXEMP.Q.28,Page.194 LA

Find the general solution of $\frac{{dy}}{{dx}} - 3y = \sin 2x$.

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

Given, $\frac{{dy}}{{dx}} - 3y = \sin 2x$

which is a linear differential equation.
On comparing it with $\frac{{dy}}{{dx}} + Py = Q,$

we get$P = - 3,Q = \sin 2x$
$IF = {e^{ - 3\int d x}} = {e^{ - 3x}}$
The general

Solution

is
$y \cdot {e^{ - 3x}} = \int {\mathop {\sin 2x}\limits_{\rm{I}} \mathop {{e^{ - 3x}}}\limits_{{\rm{II}}} } dx$

Let $\quad y \cdot {e^{ - 3x}} = I$
…….(i)
$\therefore I = \int {\mathop {{e^{ - 3x}}}\limits_{{\rm{II}}} } \mathop {\sin 2x}\limits_{\rm{I}}$

$\Rightarrow$ $I = \sin 2x\left( { \cdot \frac{{{e^{ - 3x}}}}{{ - 3}}} \right) - \int 2 \cos 2x\left( {\frac{{{e^{ - 3x}}}}{{ - 3}}} \right)dx + {C_1}$

$\Rightarrow$ $I = - \frac{1}{3}{e^{ - 3x}}\sin 2x + \frac{2}{3}\int {\mathop {e_{}^{ - 3x}}\limits_{{\rm{II}}} } \mathop {\cos 2x}\limits_{\rm{I}} dx + {C_1}$

$\Rightarrow$ $I = - \frac{1}{3}{e^{ - 3x}}\sin 2x + \frac{2}{3}\left( {\cos 2x\frac{{{e^{ - 3x}}}}{{ - 3}} - \int {( - 2\sin 2x)} \frac{{{e^{ - 3x}}}}{{ - 3}}dx} \right) + {C_1} + {C_2}$

$\Rightarrow$ $I = - \frac{1}{3}{e^{ - 3x}}\sin 2x - \frac{2}{9}\cos 2x{e^{ - 3x}} - \frac{4}{9}I + {C^\prime }$

[where,
${C^\prime } = {C_1} + {C_2}$]
$\Rightarrow$ $I + \frac{{4I}}{9}2 = + {e^{ - 3x}}\left( { - \frac{1}{3}\sin 2x - \frac{2}{9}\cos 2x} \right) + {C^\prime }$

$\Rightarrow$ $\frac{{13}}{9}I = {e^{ - 3x}}\left( { - \frac{1}{3}\sin 2x - \frac{2}{9}\cos 2x} \right) + {C^\prime }$

$\Rightarrow$ $I = \frac{9}{{13}}{e^{ - 3x}}\left( { - \frac{1}{3}\sin 2x - \frac{2}{9}\cos 2x} \right) + C$

[where $C = \frac{{9{C^\prime }}}{{13}}$]
$\Rightarrow$ $I = \frac{3}{{13}}{e^{ - 3x}}\left( { - \sin 2x - \frac{2}{3}\cos 2x} \right) + C$

$\Rightarrow$ $= \frac{3}{{13}}{e^{ - 3x}}\frac{{( - 3\sin 2x - 2\cos 2x)}}{3} + C$

$\Rightarrow$ $= \frac{{{e^{ - 3x}}}}{{13}}( - 3\sin 2x - 2\cos 2x) + C$

$\Rightarrow$ $I = - \frac{{{e^{ - 3x}}}}{{13}}(2\cos 2x + 3\sin 2x) + C$

On substituting the value of
$I$ in Eq. (i),

we get $y \cdot {e^{ - 3x}} = - \frac{{{e^{ - 3x}}}}{{13}}(2\cos 2x + 3\sin 2x) + C$

$\Rightarrow$ $y = - \frac{1}{{13}}(2\cos 2x + 3\sin 2x) + C{e^{3x}}$

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