- #1

karush

Gold Member

MHB

- 3,269

- 5

$\textsf{Find the solution of the given initial value problem. State the interval in which the solution is valid.}$

$$xy^\prime+2y=x^2-x+1, \qquad y(1)=\frac{1}{2}$$

$\textit{rewrite}$

$$y' +\frac{2}{x}y=x-1+\frac{1}{x}$$

$\textit {Find u(x) }$

$$\displaystyle u(x)=\exp\int \frac{2}{x} \, dx =\ln e^{x^2}=x^2$$

$\textit{multiply thru with $u(x)=x^2$}$

$$x^2y' +(x^2)'y=x^2(x^2-x+1)$$

$\textsf{rewrite:}$

$$(x^2 y)'=x^4-x^3+x^2 \, \color{red}{\textit{I think the problem is here}}$$

$\textit{Integrate }$

$$\displaystyle x^2y=\int x^4-x^3+x^2 \, dx

=\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}+c$$

$\textit{divide thru by $x^2$}$

$$\displaystyle y=\frac{x^3}{5}-\frac{x^2}{4}+\frac{x}{3}+\frac{c}{x^2}$$

$\textit{book answer(I couldn't get this) }$

$$y=\color{red}{\frac{1}{12}(3x^2-4x+6+\frac{1}{x^2})}$$

$$xy^\prime+2y=x^2-x+1, \qquad y(1)=\frac{1}{2}$$

$\textit{rewrite}$

$$y' +\frac{2}{x}y=x-1+\frac{1}{x}$$

$\textit {Find u(x) }$

$$\displaystyle u(x)=\exp\int \frac{2}{x} \, dx =\ln e^{x^2}=x^2$$

$\textit{multiply thru with $u(x)=x^2$}$

$$x^2y' +(x^2)'y=x^2(x^2-x+1)$$

$\textsf{rewrite:}$

$$(x^2 y)'=x^4-x^3+x^2 \, \color{red}{\textit{I think the problem is here}}$$

$\textit{Integrate }$

$$\displaystyle x^2y=\int x^4-x^3+x^2 \, dx

=\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}+c$$

$\textit{divide thru by $x^2$}$

$$\displaystyle y=\frac{x^3}{5}-\frac{x^2}{4}+\frac{x}{3}+\frac{c}{x^2}$$

$\textit{book answer(I couldn't get this) }$

$$y=\color{red}{\frac{1}{12}(3x^2-4x+6+\frac{1}{x^2})}$$

Last edited: