Finn løsningen til initialverdiproblemet
[symbol:rot] (1-x^2) dy/dx +y = x
y(0)= -1
initialverdiproblem
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Andreas345
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[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.
[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.