Question

In Problems 39-44 solve the given differential equation subject to the indicated conditions.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sinx}$

Short answer

The solution is
$\mathrm{y}=\frac{13}{4}{\mathrm{e}}^0-\frac{5}{4}{\mathrm{e}}^{-\mathrm{x}}-\mathrm{x}-\frac{1}{2}\text{sin}\mathrm{x}$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

$\mathrm{y}\text{'}\text{'}-\mathrm{y}=\mathrm{x}+\mathrm{sinx}$

Find homogeneous equation

We have the second order differential equation

$y^{\text{'}\text{'}}-y=x+\text{sin}x$

as the form

$y^{\text{'}\text{'}}+p\left(x\right)y^{\text{'}}+q\left(x\right)y=f\left(x\right)$

with the conditions

role="math" localid="1667894358983" $y\left(0\right)=2and{y}^{\text{'}}\left(0\right)=3,$

and we have to obtain its general solution as the following technique:
First, we have to obtain the solution of the corresponding homogeneous D.E$y\text{'}\text{'}-2y^{\text{'}}+y=0$for the given D.E by assuming that$y=e^{mx}$is a solution of the given D.E, then follow this technique:
Differentiate this assumption with respect to$x$, then we have

$\begin{array}{cc}{y}^{\text{'}}& =m{e}^{mx}\\ y^{\text{'}\text{'}}& =m^2{e}^{mx}\end{array}$

Find auxiliary equation and general solution

After that, substitute with $y\text{'}$and $y\text{'}\text{'}$shown in equations $\left(2\right)$ and $\left(3\right)$ and our assumption $y=e^{mx}$into equation , then we obtain

$\begin{array}{rr}{m}^2{e}^{mx}-e^{mx}& =0\\ (m^2-1)e^{mx}& =0\end{array}$

Since$e^{mx}$cannot be 0, then we have the auxiliary equation as

$\begin{array}{c}{m}^2-1=0\\ (m-1)(m+1)=0\end{array}$

Then we have its roots as

$m_1=1\text{and}m=-1$

which are real and distinct.
Then we can obtain the corresponding homogeneous solution of the given D.E as

$y_c=c_1{e}^x+c_2{e}^{-x}$

Find particular solution

Second, we have to obtain the particular solution of the given differential equation by assuming that $y_p=Ax+B\cos x+C\sin x$, since we have the free term of the given D.E as $g\left(x\right)=x+\text{sin}x$, then differentiate with respect to $x$as

$\begin{array}{c}{y}_p^{\text{'}}=A-B\text{sin}x+C\text{cos}x\\ y_p^{\text{''}}=-B\text{cos}x-C\text{sin}x\end{array}$

After that, substitute with$y_p^{\text{'}\text{'}}$shown in equation$\left(4\right)$and our assumption$y_p=Ax+B\cos x+C\sin x$into equation (1), then we obtain

$(-B\text{cos}x-C\text{sin}x)-(Ax+B\text{cos}x+C\text{sin}x)=x+\text{sin}x$

$-Ax+(-2B)\text{cos}x+(-2C)\text{sin}x=x+\text{sin}x$

Then by comparison we have

$\begin{array}{c}-A=1\text{then}:A=-1\\ \text{and}-2B=0\text{then}:B=0\\ \text{and}-2C=1\text{then}:C=-\frac{1}{2}\end{array}$

Then we can have the particular solution as

$y_p=-x-\frac{1}{2}\text{sin}x$

Then from equations$\left(a\right)$and$\left(b\right)$, we can obtain the general solution of the given D.E as

$\begin{array}{rl}y& =y_c+y_p\\ & =c_1{e}^x+c_2{e}^{-x}-x-\frac{1}{2}\text{sin}x\end{array}$

Final Answer

Third, we have to apply the given conditions into the solution of the given differential equation to find the constants$c_1$and$c_2$as the following:
Apply the point$(x,y)=(0,2)$into equation$\left(5\right)$, then we have

$\begin{array}{c}2=c_1{e}^0+c_2{e}^0-0-\frac{1}{2}\text{sin}\left(0\right)\\ c_1+c_2=2\end{array}$

Second, we have to take the first derivative for the solution of the given differential equation in (4) as

$y^{\text{'}}=c_1{e}^x-c_2{e}^{-x}-1-\frac{1}{2}\text{cos}x$

Then apply the point $(x,y^{\text{'}})=(0,3)$into equation $\left(5\right)$, then we have

$\begin{array}{c}3=c_1{e}^0-c_2{e}^0-1-\frac{1}{2}\text{cos}\left(0\right)\\ 3=c_1-c_2-\frac{3}{2}\\ c_1-c_2=\frac{9}{2}\end{array}$

By solving equations$\left(c\right)$and$\left(d\right)$ , we find that

$c_1=\frac{13}{4}\text{and}{c}_2=-\frac{5}{4}$

Finally, substitute with the constants$c_1$and$c_2$into equation, then we obtain

$y=\frac{13}{4}{e}^0-\frac{5}{4}{e}^{-x}-x-\frac{1}{2}\text{sin}x$

is the solution for the given differential equation at the given conditions.

Thus, the required solution is

$y=\frac{13}{4}{e}^0-\frac{5}{4}{e}^{-x}-x-\frac{1}{2}\text{sin}x$