Question

Solve the given initial value problem. Give the largest interval l over which the general solution is defined.

$\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{xy}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

The solution for the given initial value is $y=\frac{1}{6}{x}^2{e}^{x^2}-\frac{1}{18}{e}^{x^2}-\frac{17}{18}{e}^{-2x^2}$ .

Step-by-step solution

The given equation in the standard form and determine the integrated factor

Consider the following initial value problem:

$y\text{'}+4xy=x^3{e}^{x^2},y\left(0\right)=-1$

The objective is to solve the following initial value problem (IVP) and give the largest interval over which the solution is defined. Rewrite the given differential equation as,

$\frac{dy}{dx}+4xy=x^3{e}^{x^2}$ …….. (1)

Compare the given differential equation with the linear equation of the form

$$\begin{gathered} \frac{dy}{dx}+P\left(x\right)y=Q\left(x\right) \\ P\left(x\right)=4x \end{gathered}$$

The integrating factor is,

$$\begin{gathered} e^{\int p\left(x\right)dx}=e^{\int 4xdx} \\ =e^{4\int xdx} \\ =e^{2x^2} \end{gathered}$$

Thus, the integrating factor is,

$e^{\int p\left(x\right)dx}=e^{2x^2}$

Determine the general solution for the given differential equation

Multiply the differential equation (1) with integrating factor $e^{\int p\left(x\right)dx}=e^{2x^2}$

$\Rightarrow e^{2x^2}\left(\frac{dy}{dx}+4xy\right)=x^3{e}^{x^2}$

Now, integrate on both sides and solve for x .

$$\begin{gathered} \Rightarrow e^{2x^2}\left(\frac{dy}{dx}+4xy\right)=x^3{e}^{x^2} \\ \Rightarrow \int d\left(e^{2x^2}y\right)=\int x^3{e}^{3x^2}dx \\ \Rightarrow y{e}^{2x^2}=\frac{1}{6}{x}^2{e}^{3x^2}-\frac{1}{18}{e}^{3x^2}+C \end{gathered}$$

Therefore, the general solution of the differential equation is $y{e}^{2x^2}=\frac{1}{6}{x}^2{e}^{3x^2}-\frac{1}{18}{e}^{3x^2}+C$.

$$\begin{gathered} y{e}^{2x^2}=\frac{1}{6}{x}^2{e}^{3x^2}-\frac{1}{18}{e}^{3x^2}+C\backslash \\ \Rightarrow (-1)e^0=\frac{1}{6}\left(0\right)e^0-\frac{1}{18}{e}^0+C \\ y{e}^{2x^2}=\frac{1}{6}{x}^2{e}^{3x^2}-\frac{1}{18}{e}^{3x^2}+C \\ y{e}^{2x^2}=\frac{1}{6}{x}^2{e}^{3x^2}-\frac{1}{18}{e}^{3x^2}-\frac{17}{18} \\ y=\frac{\frac{1}{6}{x}^2{e}^{3x^2}-\frac{1}{18}{e}^{3x^2}-\frac{17}{18}}{e^{2x^2}} \\ y=\frac{1}{6}{x}^2{e}^{x^2}-\frac{1}{18}{e}^{x^2}-\frac{17}{18}{e}^{-2x^2} \end{gathered}$$

Therefore, the solution to the initial value problem (1) is $y=\frac{1}{6}{x}^2{e}^{x^2}-\frac{1}{18}{e}^{x^2}-\frac{17}{18}{e}^{-2x^2}$

The general solution $y=\frac{1}{6}{x}^2{e}^{x^2}-\frac{1}{18}{e}^{x^2}-\frac{17}{18}{e}^{-2x^2}$is a polynomial, so it defined for all real numbers Hence, the largest interval in which the solution defined is role="math" $-\infty <y<\infty$.