Question

In Problems express the solution of the given initial-value problem in terms of an integraldefined function.

$\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{x}{\mathit{e}}^{{\mathbf{x}}^{\mathbf{2}}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Short answer

The solution is $y\left(x\right)=x^{-2}{\int }_1^x{t}^2{e}^{t^2}\text{d}t+3x^{-2}$.

Step-by-step solution

Definition

The standard form of a linear differential equation is $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{Py}\mathbf{+}\mathbf{Q}$, and it contains the variabley, and its derivatives.

Find integrating factor

By the general form of linear differential equation, we obtain

$P\left(x\right)=\frac{2}{x}\text{and}f\left(x\right)=e^{x^2}$

The integrating factor is given by$e^{\int P\left(x\right)dx}$

The integrating factor is:

$\begin{array}{rcl}{e}^{\int P\left(x\right)\text{d}x}& =& e^{2\int \frac{1}{x}\text{d}x}\\ & =& e^{2\ln x}\\ & =& e^{\ln x^2}\\ & =& x^2\end{array}$

Solution

Thus, the solution is:

$\begin{array}{rcl}y{x}^2& =& {\int }_1^x{t}^2{e}^{t^2}\text{d}t+c\\ y& =& x^{-2}{\int }_1^x{t}^2{e}^{t^2}\text{d}t+c{x}^{-2} -----\left(1\right)\end{array}$

Apply initial condition

Use the initial condition $y\left(1\right)=3$, to get .

Substitute $c=3$in the equation (1).

$y\left(x\right)=x^{-2}{\int }_1^x{t}^2{e}^{t^2}\text{d}t+3x^{-2}$

So, the solution of given differential equation is$y\left(x\right)=x^{-2}{\int }_1^x{t}^2{e}^{t^2}\text{d}t+3x^{-2}$