8ty+e^ty^'= y/(t^2+64)

Lösung

8 t y + e^{t} y^{^{'}} = \frac{y}{t ^{2} + 64}

y = e^{\int \frac{1 - 8 t ( t ^{2} + 64 )}{e ^{t} ( t ^{2} + 64 )} d t}

Schritte zur Lösung

8 t y + e^{t} y^{'} (t) = \frac{y}{t ^{2} + 64}

Solve separable ODE:

y = e^{\int \frac{1 - 8 t ( t ^{2} + 64 )}{e ^{t} ( t ^{2} + 64 )} d t}

y = e^{\int \frac{1 - 8 t ( t ^{2} + 64 )}{e ^{t} ( t ^{2} + 64 )} d t}

y y^{^{'}} - 4 y^{^{'}} - 5 y = 125

(x + 1) (x - 2) \frac{d y}{d x} + y = 0

3 (16 + y^{2}) d x + 2 x y d y = 0

(y + x y) + (x - x y) y^{^{'}} = 0

\frac{d y}{d x} + y = y x e^{x + 8}