Question

Solve $\left( {{D}^{2}}+4D+3 \right)y={{e}^{-x}}\sin x+x{{e}^{3x}}$?

Answer 1

First of all find the complementary function (C.F.) by equating $\left( {{D}^{2}}+4D+3 \right)=0$ and solving for the value of D. Write the complementary function as $A{{e}^{-{{m}_{1}}x}}+B{{e}^{-{{m}_{2}}x}}$ where A and B are the constants and ${{m}_{1}}$, ${{m}_{2}}$ are the roots of the equation $\left( {{D}^{2}}+4D+3 \right)=0$. Now, find the particular integral (P.I.) by writing the differential equation as $y=\dfrac{1}{\left( {{D}^{2}}+4D+3 \right)}\left( {{e}^{-x}}\sin x+x{{e}^{3x}} \right)$. Write $\dfrac{1}{\left( {{D}^{2}}+4D+3 \right)}$ as the sum of difference of two linear terms using the partial fraction and solve the R.H.S by using the formula $\left( \dfrac{1}{D-a} \right)X={{e}^{ax}}\int{X{{e}^{-ax}}dx}$ where X is the given functions of x. Use the ILATE rule to find the integral product of two functions.

Complete step-by-step answer:
Here we have been provided with the second order linear differential equation $\left( {{D}^{2}}+4D+3 \right)y={{e}^{-x}}\sin x+x{{e}^{3x}}$ and we are asked to solve it.
Now, the solution of a second order linear differential equation contains two parts. First part is known as complementary function (C.F.) and the second part is known as the particular integral. Let us find them one by one.
(1) To find the C.F. of the differential equation we have to substitute $f\left( D \right)=\left( {{D}^{2}}+4D+3 \right)=0$, so on solving the quadratic equation in D using the middle term split method we get,
$\begin{aligned} & \Rightarrow \left( {{D}^{2}}+4D+3 \right)=0 \\\ & \Rightarrow \left( D+1 \right)\left( D+3 \right)=0 \\\ \end{aligned}$
Substituting each term equal to 0 one by one we get,
$\Rightarrow D=-1$ or $D=-3$
Therefore, the complementary function is given as: -
$\Rightarrow C.F.=A{{e}^{-x}}+B{{e}^{-3x}}$, where (A) and (B) are constants.
(2) Now, to find the particular integral we follow the following procedure.
$\Rightarrow y=\dfrac{1}{\left( {{D}^{2}}+4D+3 \right)}\left( {{e}^{-x}}\sin x+x{{e}^{3x}} \right)$
Writing $\dfrac{1}{\left( {{D}^{2}}+4D+3 \right)}$ as the sum of difference of two linear terms using the partial fraction we get,
$\begin{aligned} & \Rightarrow y=\dfrac{1}{2}\left( \dfrac{1}{D+1}-\dfrac{1}{D+3} \right)\left( {{e}^{-x}}\sin x+x{{e}^{3x}} \right) \\\ & \Rightarrow y=\dfrac{1}{2}\left( \dfrac{1}{D+1}\left( {{e}^{-x}}\sin x \right)+\dfrac{1}{D+1}\left( x{{e}^{3x}} \right)-\left( \dfrac{1}{D+3}\left( {{e}^{-x}}\sin x \right)+\dfrac{1}{D+3}\left( x{{e}^{3x}} \right) \right) \right) \\\ \end{aligned}$
Let us find the values of each operation one by one. Here we have to use the formula $\left( \dfrac{1}{D-a} \right)X={{e}^{ax}}\int{X{{e}^{-ax}}dx}$, where X is the given functions of x, to solve for the different expressions.
(i) Considering the expression $\dfrac{1}{D+1}\left( {{e}^{-x}}\sin x \right)$ we have,
$\begin{aligned} & \Rightarrow \dfrac{1}{D+1}\left( {{e}^{-x}}\sin x \right)={{e}^{-x}}\int{{{e}^{-x}}\sin x\times {{e}^{x}}dx} \\\ & \Rightarrow \dfrac{1}{D+1}\left( {{e}^{-x}}\sin x \right)={{e}^{-x}}\int{\sin xdx} \\\ & \Rightarrow \dfrac{1}{D+1}\left( {{e}^{-x}}\sin x \right)=-{{e}^{-x}}\cos x \\\ \end{aligned}$
(ii) Considering the expression $\dfrac{1}{D+1}\left( x{{e}^{3x}} \right)$ we have,
$\begin{aligned} & \Rightarrow \dfrac{1}{D+1}\left( x{{e}^{3x}} \right)={{e}^{-x}}\int{x{{e}^{3x}}\times {{e}^{x}}dx} \\\ & \Rightarrow \dfrac{1}{D+1}\left( x{{e}^{3x}} \right)={{e}^{-x}}\int{x{{e}^{4x}}dx} \\\ \end{aligned}$
Using the ILATE rule for the integration we get,
$\begin{aligned} & \Rightarrow \dfrac{1}{D+1}\left( x{{e}^{3x}} \right)={{e}^{-x}}\left[ \dfrac{x{{e}^{4x}}}{4}-\dfrac{1}{4}\int{{{e}^{4x}}dx} \right] \\\ & \Rightarrow \dfrac{1}{D+1}\left( x{{e}^{3x}} \right)={{e}^{-x}}\left[ \dfrac{x{{e}^{4x}}}{4}-\dfrac{{{e}^{4x}}}{16} \right] \\\ & \Rightarrow \dfrac{1}{D+1}\left( x{{e}^{3x}} \right)={{e}^{3x}}\left[ \dfrac{x}{4}-\dfrac{1}{16} \right] \\\ \end{aligned}$
(iii) Considering the expression $\dfrac{1}{D+3}\left( {{e}^{-x}}\sin x \right)$ we have,
$\begin{aligned} & \Rightarrow \dfrac{1}{D+3}\left( {{e}^{-x}}\sin x \right)={{e}^{-3x}}\int{{{e}^{-x}}\sin x\times {{e}^{3x}}dx} \\\ & \Rightarrow \dfrac{1}{D+3}\left( {{e}^{-x}}\sin x \right)={{e}^{-3x}}\int{{{e}^{2x}}\sin xdx} \\\ \end{aligned}$
Assuming $\int{{{e}^{2x}}\sin xdx}$ as I and using the ILATE rule for the integration we get,
$\begin{aligned} & \Rightarrow I=\int{{{e}^{2x}}\sin xdx} \\\ & \Rightarrow I=\sin x\times \dfrac{{{e}^{2x}}}{2}-\int{\dfrac{{{e}^{2x}}\cos x}{2}dx} \\\ & \Rightarrow I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{1}{2}\int{{{e}^{2x}}\cos xdx} \\\ \end{aligned}$
Again using the ILATE rule we get,

& \Rightarrow I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{1}{2}\left[ \cos x\times \dfrac{{{e}^{2x}}}{2}-\int{-\dfrac{{{e}^{2x}}\sin x}{2}dx} \right] \\\ & \Rightarrow I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{1}{2}\left[ \dfrac{{{e}^{2x}}\cos x}{2}+\dfrac{1}{2}\int{{{e}^{2x}}\sin xdx} \right] \\\ & \Rightarrow I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{1}{2}\left[ \dfrac{{{e}^{2x}}\cos x}{2}+\dfrac{I}{2} \right] \\\ \end{aligned}$$ Solving for the value of I we get, $$\Rightarrow I=\dfrac{{{e}^{2x}}\left( 2\sin x-\cos x \right)}{5}$$ $\Rightarrow \dfrac{1}{D+3}\left( {{e}^{-x}}\sin x \right)=\dfrac{{{e}^{-x}}}{5}\left( 2\sin x-\cos x \right)$ (iv) ) Considering the expression $\dfrac{1}{D+3}\left( x{{e}^{3x}} \right)$ we have, $\begin{aligned} & \Rightarrow \dfrac{1}{D+3}\left( x{{e}^{3x}} \right)={{e}^{-3x}}\int{x{{e}^{3x}}\times {{e}^{3x}}dx} \\\ & \Rightarrow \dfrac{1}{D+3}\left( x{{e}^{3x}} \right)={{e}^{-3x}}\int{x{{e}^{6x}}dx} \\\ \end{aligned}$ Using the ILATE rule for the integration we get, $\begin{aligned} & \Rightarrow \dfrac{1}{D+3}\left( x{{e}^{3x}} \right)={{e}^{-3x}}\left[ \dfrac{x{{e}^{6x}}}{6}-\dfrac{1}{6}\int{{{e}^{6x}}dx} \right] \\\ & \Rightarrow \dfrac{1}{D+3}\left( x{{e}^{3x}} \right)={{e}^{-3x}}\left[ \dfrac{x{{e}^{6x}}}{6}-\dfrac{{{e}^{6x}}}{36} \right] \\\ & \Rightarrow \dfrac{1}{D+3}\left( x{{e}^{3x}} \right)={{e}^{3x}}\left[ \dfrac{x}{4}-\dfrac{1}{16} \right] \\\ \end{aligned}$ Therefore, the particular integral can be given as: - $$\Rightarrow P.I.=\dfrac{1}{2}\left( -{{e}^{-x}}\cos x+{{e}^{3x}}\left( \dfrac{x}{4}-\dfrac{1}{16} \right)-\left( \dfrac{{{e}^{-x}}}{5}\left( 2\sin x-\cos x \right)+{{e}^{3x}}\left( \dfrac{x}{6}-\dfrac{1}{36} \right) \right) \right)$$ On simplification we get, $$\begin{aligned} & \Rightarrow P.I.=\dfrac{1}{2}\left( -\dfrac{{{e}^{-x}}}{5}\left( 4\cos x+2\sin x \right)+\dfrac{{{e}^{3x}}}{144}\left( 12x-5 \right) \right) \\\ & \Rightarrow P.I.=\left( -\dfrac{{{e}^{-x}}}{5}\left( 2\cos x+\sin x \right)+\dfrac{{{e}^{3x}}}{288}\left( 12x-5 \right) \right) \\\ \end{aligned}$$ Hence, the solution of the given differential equation is: - $$\begin{aligned} & \Rightarrow y=P.I.+C.F. \\\ & \therefore y=A{{e}^{-x}}+B{{e}^{-3x}}-\dfrac{{{e}^{-x}}}{5}\left( 2\cos x+\sin x \right)+\dfrac{{{e}^{3x}}}{288}\left( 12x-5 \right) \\\ \end{aligned}$$ **Note:** Note that the given differential equation is of second order which we generally study in higher mathematics. You can also be provided with differential equations of order 3 or 4 and in such cases also you have to follow the same procedure of calculating the C.F. and P.I. Remember the basic formula $\left( \dfrac{1}{D-a} \right)X={{e}^{ax}}\int{X{{e}^{-ax}}dx}$ as it will be used in cases where you may forget some special formulas.