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Finn løsningen til initialverdiproblemet

[symbol:rot] (1-x^2) dy/dx +y = x

y(0)= -1

[tex]\sqrt{1-x^2}\cdot \frac{dy}{dx}+y=x[/tex]

[tex]\frac{dy}{dx}+\frac{1}{sqrt{1-x^2}}\cdot y=\frac{x}{sqrt{1-x^2}}[/tex]

Integrerende faktor: [tex]e^{\int \frac{1}{sqrt{1-x^2}}}=e^{asin(x)}[/tex]

[tex]\frac{dy}{dx}\cdot e^{asin(x)}+\frac{1}{sqrt{1-x^2}}\cdot y\cdot e^{asin(x)}=\frac{x}{sqrt{1-x^2}}\cdot e^{asin(x)}[/tex]

[tex]\left (e^{asin(x)}\cdot y \right )\prime=\frac{x}{sqrt{1-x^2}}\cdot e^{asin(x)}[/tex]


[tex]e^{asin(x)}\cdot y=\int \frac{x}{sqrt{1-x^2}}\cdot e^{asin(x)}\ dx [/tex]

Tips fra her: [tex]u=asin(x) \Leftrightarrow x=sin(u)[/tex] og [tex]du=\frac{1}{sqrt{1-x^2}} dx[/tex]

Benytt så delvis integrasjon.