Hi,
ich habe gegeben:
$$ y' = xy + \frac { x }{ y } \quad ; \quad y(0) = 1 \quad (x \geq 0)$$
Wie löse ich es?
Meine Rechnungen bisher:
$$ \frac { dy }{ dx } = xy + \frac { x }{ y } $$
$$ dy = xydx + \frac { x }{ y } dx $$
$$ \frac { 1 }{ y }dy = xdx + x\frac { 1 }{ y^2 }dx $$
$$ \frac { 1 }{ y }dy = xdx(\frac { y^2+1 }{ y^2 }) $$
$$ \frac { y^2 }{ y\cdot (y^2+1) }dy = xdx $$
$$ \frac { y^2 }{ y^2\cdot (y+\frac { 1 } { y^2 }) }dy = xdx $$
$$ \frac { 1 }{ y+ \frac { 1 }{ y^2 } }dy = xdx $$
$$ \int_{}^{} \frac { 1 }{ y+ \frac { 1 }{ y^2 } }dy = \int_{}^{} xdx $$
$$ \frac { 1 }{ 3 } ln(|y^3+1|) = \frac { 1 }{ 2 } x^2 + c $$
$$ e^{ln(|y^3+1|)} = e^{\frac { 1 }{ 2 } x^2 + c} $$
$$ |y^3+1| = e^{\frac { 1 }{ 2 }x^2}e^c $$
$$ y^3+1 = ±e^c e^{\frac { 1 }{ 2 }x^2} $$
$$ y^3 = ±e^c e^{\frac { 1 }{ 2 }x^2} - 1 \quad ; \quad A = ±e^c$$
$$ y^3 = A e^{\frac { 1 }{ 2 }x^2} - 1 $$
$$ y = \sqrt [ 3 ]{ A e^{\frac { 1 }{ 2 }x^2} - 1 } $$
Irgendwo ein Fehler bis jetzt?
