Question

Proceed as in Example 7 and express the solution of the given initial value problem in terms of an integral defined function.

${\mathbf{x}}^{\mathbf{2}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

So, the solution of the given initial value problem in terms of an integral defined function $y=e^{-\frac{1}{x}\left[{\int }_0^{x}t{e}^{\frac{1}{x}}\mathrm{d}t-{\int }_0^{1}t{e}^{\frac{1}{x}}\mathrm{d}t\right]}$.

Step-by-step solution

Form of linear equation

The linear differential equation is of the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{Py}\mathbf{=}\mathbf{Q}$, where P and Q are numeric constants or functions in x.

Evaluation

The given differential equation, can be written as

$\frac{\mathrm{d}y}{\mathrm{d}x}-y{x}^{-2}=x$

Linear differential equation of the first order

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

The integrating factor is given by

$e^{\int P\left(x\right)dx}$

Based on (1) and the given differential equation, we obtain

$P\left(x\right)=-x^{-2}\mathrm{and}f\left(x\right)=x$

Find integrated factor

Use the formula (2), to get integrating factor

$\begin{array}{l}{e}^{\int P\left(x\right)dx}=e^{-\int x^{-2}dx}\\ =e^{\frac{1}{x}}\end{array}$

Thus, the solution is $y{e}^{\frac{1}{x}}={\int }_0^{x}t{e}^{\frac{1}{1}}\mathrm{d}t+c$.

Find the value of constant

Use the initial condition $y\left(1\right)=0$to get c

$\begin{array}{l}0={\int }_0^{1}t{e}^{\frac{1}{2}}\mathrm{d}t+c\\ c=-{\int }_0^{1}t{e}^{\frac{1}{t}}\mathrm{d}t\end{array}$

Substitute the value $c=-{\int }_0^{1}t{e}^{\frac{1}{2}}\mathrm{d}t$ in the equation (3)

$\begin{array}{l}y{e}^{\frac{1}{x}}={\int }_0^{x}t{e}^{\frac{1}{x}}\mathrm{d}t-{\int }_0^{1}t{e}^{\frac{1}{t}}\mathrm{d}t\\ y=e^{-\frac{1}{x}\left[{\int }_0^{x}t{e}^{\frac{1}{x}}\mathrm{d}t-{\int }_0^{1}t{e}^{\frac{1}{x}}\mathrm{d}t\right]}\end{array}$

So, the required solution is $y=e^{-\frac{1}{x}\left[{\int }_0^{x}t{e}^{\frac{1}{x}}\mathrm{d}t-{\int }_0^{1}t{e}^{\frac{1}{x}}\mathrm{d}t\right]}$.