4xy^'-20y=x^{-9}

Lösung

4 x y^{^{'}} - 20 y = x^{- 9}

y = - \frac{1}{56 x ^{9}} + c_{1} x^{5}

Schritte zur Lösung

4 x y^{'} (x) - 20 y = x^{- 9}

Solve linear ODE:

y = - \frac{1}{56 x ^{9}} + c_{1} x^{5}

y = - \frac{1}{56 x ^{9}} + c_{1} x^{5}

\int_{0}^{π} x \cdot sin (x) d x

\int \frac{2 x ^{2} + 3}{( x ^{2} + 1 ) ^{2}} d x

\int_{- 1}^{4} (x^{2} + 2 x - 1) d x

\frac{d}{d x} (- y x)

49 y^{^{''}} + 140 y^{^{'}} + 100 y = 0