Question

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

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{4}\mathbf{xy}\mathbf{=}{\mathbf{sinx}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{7}$

Short answer

The solution is$y\left(x\right)=e^{2x^2}{\int }_0^x{e}^{-2t^2}\sin t^2\text{d}t+7e^{2x^2}$

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

$\begin{array}{rcl}P\left(x\right)& =& -4x\text{and}f\left(x\right)\\ & =& \sin x^2\end{array}$

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

Use the formula the integrating factor is:

$\begin{array}{rcl}{e}^{\int P\left(x\right)\text{d}x}& =& e^{-4\int x\text{d}x}\\ & =& e^{-2x^2}\end{array}$

Solution

The solution is thus given by:

$$\begin{gathered} y{e}^{-2x^2}={\int }_0^x{e}^{-2t^2}\sin t^2\text{d}t+c \\ y=e^{2x^2}{\int }_0^x{e}^{-2t^2}\sin t^2\text{d}t+c{e}^{2x^2} -----\left(1\right) \end{gathered}$$

Apply initial condition

Use the initial condition $y\left(0\right)=7$, to get c.

Substitute the value $c=7$in the equation (1).

$y\left(x\right)=e^{2x^2}{\int }_0^x{e}^{-2t^2}\sin t^2\text{d}t+7e^{2x^2}$

So, the solution of $\frac{dy}{dx}-4xy=\sin x^2, y\left(0\right)=7$is$y\left(x\right)=e^{2x^2}{\int }_0^x{e}^{-2t^2}\sin t^2\text{d}t+7e^{2x^2}$