Question

Solve:

$\frac{dy}{dx}+2y=x{e}^{4x}$.

Answer 1

Given differential eqn is:-

$\frac{dy}{dx}+2y=x{e}^{4x}$

Clearly, It is as linear differential eqn of the form $\frac{dy}{dx}+Py=Q$ where $P=2,Q=x{e}^{4x}$

Integrating factor I.E. is given by

$I.E.=e^{\int Pdx}=e^{\int 2dx}=e^{2x}$

we know solutions of linear differential eqn is given by

$y\times I.F.=\int (I.F\times Q)dx+C$

$\Rightarrow y\times e^{2x}=\int x{e}^{4x}.e^{2x}dx+c$

$\Rightarrow y\times e^{2x}=\int x{e}^{6x}dx+c(\because a^x.a^y=a^{x+y})$

$\Rightarrow y\times e^{2x}=x.\int e^{6x}dx-\int (\frac{dx}{dx}.\int e^{6x}dx)dx+c$ ( Using ILATE rule)

$\Rightarrow y\times e^{2x}=x\frac{e^{6x}}{6}-\int 1.\frac{e^{6x}}{6}dx+c$ $\left(\int u.vdx=u.\int vdx-\int (\frac{dv}{dx}.\int vdv)dx\right)$

$\Rightarrow y\times e^{2x}=x\frac{e^{6x}}{6}-\frac{e^{6x}}{36}+c$

$\therefore y=\frac{x{e}^{4x}}{6}-\frac{e^{4x}}{4}+c{e}^{-2x}$

final Ans: $\to y=e^{4x}(\frac{x}{6}-\frac{1}{4}+c{e}^{-6x})$