y^'+4xy=x^3e^{x^2}

Lösung

y^{^{'}} + 4 x y = x^{3} e^{x^{2}}

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} + \frac{c _{1}}{e ^{2 x^{2}}}

Schritte zur Lösung

y^{'} (x) + 4 x y = x^{3} e^{x^{2}}

Solve linear ODE:

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} + \frac{c _{1}}{e ^{2 x^{2}}}

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} + \frac{c _{1}}{e ^{2 x^{2}}}

d x = 4 (x^{2} + 1) d t

in v erse l a pl a ce \frac{1}{( s + 5 )}

in v erse l a pl a ce \frac{1}{5 s - 2}

\int \frac{x ^{2}}{2 x} d x

\int c^{x} d x