xy^'-y= 5/2 xln(x)

Lösung

x y^{^{'}} - y = \frac{5}{2} x ln (x)

y = \frac{5}{4} ln^{2} (x) x + c_{1} x

Schritte zur Lösung

x y^{'} (x) - y = \frac{5}{2} x ln (x)

Solve linear ODE:

y = \frac{5}{4} ln^{2} (x) x + c_{1} x

y = \frac{5}{4} ln^{2} (x) x + c_{1} x

im pl i c i t \frac{d y}{d x}, 3 e^{x y} - x = 0

\int z^{2} d z

\frac{\partial}{\partial t} ((x - c t)^{4} + (x + c t)^{4})

\int \frac{( 4 x + 2 )}{x ^{3} - x ^{2} + 2 x} d x

y^{^{'}} = 2 t y^{2}