5x^2y+4xy^'+3y=0

Lösung

5 x^{2} y + 4 x y^{^{'}} + 3 y = 0

y = \frac{e ^{\frac{- 5 x ^{2} + c _{1}}{8}}}{x ^{\frac{3}{4}}}

Schritte zur Lösung

5 x^{2} y + 4 x y^{'} (x) + 3 y = 0

Solve separable ODE:

y = \frac{e ^{\frac{- 5 x ^{2} + c _{1}}{8}}}{x ^{\frac{3}{4}}}

y = \frac{e ^{\frac{- 5 x ^{2} + c _{1}}{8}}}{x ^{\frac{3}{4}}}

y^{^{'}} = t y^{2}, y (0) = 2

\frac{d y}{d t} = 0.03 (291 - y)

\frac{d y}{d x} = \frac{cos ( 5 x )}{4 y ^{3}}

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

\frac{d x}{d y} = \frac{( x + 2 ) ^{2}}{x}