Question

Proceed as in Example 3 and solve each differential equation by variation of parameters.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{e}}^{{\mathbf{x}}^{\mathbf{2}}}$

Short answer

The solution of a given differential equation by variation of parameters is$$\begin{gathered} y=c_1\cos x+c_2\sin x \\ -\cos x\backslash {\mathrm{smallint}}_{x_0}^x{e}^{t^2}\sin tdt \\ +\sin x\backslash {\mathrm{smallint}}_{x_0}^x{e}^{t^2}\cos tdt \end{gathered}$$

Step-by-step solution

Define wronskian:

The wronskian is used to determine the solution of two functions. It can found by using the determinant,$$\begin{gathered} \mathit{W}\mathbf{(}{\mathbf{y}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\left|y_1{y}_2 \\ {y}_1\text{'}y{\text{'}}_2\right| \end{gathered}$$

Find the solution by using the required technique: 

Consider the given second order differential equation, $y\text{'}\text{'}+y=e^{x^2}$and it can be written in the form of,

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

Hence, find the solution for the differential equation as,

$y\text{'}\text{'}+y=0$ …(1)

Assume $y=e^{mx}$is a solution for the given homogeneous differential equation,

Thus, differentiate $y=e^{mx}$with respect to

$$\begin{gathered} y\text{'}=m{e}^{mt} \\ andy\text{'}\text{'}=m^2{e}^{mt} \end{gathered}$$ …(2

Substitute equation (2) in (1) with $y=e^{mt}$.

$$\begin{gathered} m^2{e}^{mt}+e^{mt}=0 \\ {e}^{mt}\left(m^2+1\right)=0 \end{gathered}$$

Thus, in the above equation e^mtcannot be equal to 0, we have

$m^2+1=0$

Hence, by simplifying we get the roots,

$m_{1,2}=\pm i$

Therefore, the solution for the homogeneous differential equation can be obtained as,

$y=c_1\cos x+c_2\sin x$

Find the solution yp by using Wronskian:

Obtain wronskian by substituting $y_1=\cos xand{y}_2=\sin x$,

$$\begin{gathered} W\left(y_1,y_2\right)=\left|y_1{y}_2 \\ {y}_1\text{'}y{\text{'}}_2\right| \\ =\left|\cos x\sin x \\ -\sin x\cos x\right| \\ ={\cos }^{2}x+{\sin }^{2}x \\ =1 \end{gathered}$$

Hence, we found wronskian W₁ and W₂, we need to find ,

$$\begin{gathered} u_1^{\text{'}}=\frac{W_1}{W\left(y_1,y_2\right)} \\ =\frac{-e^{x^2}\sin x}{1} \\ =-e^{x^2}\sin x \\ {u}_2^{\text{'}}=\frac{W_2}{W\left(y_1,y_2\right)} \\ =\frac{e^{x^2}\cos x}{1} \\ =e^{x^2}\cos x \end{gathered}$$

Substitute the obtained values and determine the solution:

Thus, by using Exampleobtain the values of u₁ and u₂,

$$\begin{gathered} u_1=\int u_1^{\text{'}}\left(t\right) \\ ={\int }_{x_0}^x-e^{t^2}\sin tdt \\ =-{\int }_{x_0}^x{e}^{t^2}\sin tdt \\ {u}_2=\int u_1^{\text{'}}\left(t\right) \\ ={\int }_{x_0}^x{e}^{t^2}\cos tdt \end{gathered}$$

Substitute the values of u₁ and u₂to obtain the solution as,

$$\begin{gathered} =\left(-{\int }_{x_0}^x{e}^{t^2}\sin tdt\right)\left(\cos x\right)+\left({\int }_{x_0}^x{e}^{t^2}\cos tdt\right)\left(\sin x\right) \\ =-\cos x{\int }_{x_0}^x{e}^{t^2}\sin tdt+\sin x{\int }_{x_0}^x{e}^{t^2}\cos tdt \end{gathered}$$

Hence, the general solution for the given differential equation can be obtained as,

$$\begin{gathered} y=y_h+y_p \\ \cos x+c_2\sin x-\cos x{\int }_{x_0}^x{e}^{t^2}\sin tdt+\sin x{\int }_{x_0}^x{e}^{t^2}\cos tdt \end{gathered}$$

Therefore, the required solution for the given differential equation is.

$y=c_1\cos x+c_2\sin x-\cos x{\int }_{x_0}^x{e}^{t^2}\sin tdt+\sin x{\int }_{x_0}^x{e}^{t^2}\cos tdt$