(x-4)y^'+y=x^2+8

Lösung

(x - 4) y^{^{'}} + y = x^{2} + 8

y = \frac{x ^{3}}{3 ( x - 4 )} + \frac{8 x}{x - 4} + \frac{c _{1}}{x - 4}

Schritte zur Lösung

(x - 4) y^{'} (x) + y = x^{2} + 8

Solve linear ODE:

y = \frac{x ^{3}}{3 ( x - 4 )} + \frac{8 x}{x - 4} + \frac{c _{1}}{x - 4}

y = \frac{x ^{3}}{3 ( x - 4 )} + \frac{8 x}{x - 4} + \frac{c _{1}}{x - 4}

d er i v a t i v e 3 x^{3}

\frac{d}{d y} (ln (y))

\int x sin (x^{2} + 4) d x

x \frac{d y}{d x} - 2 y = 6 x^{5}

y^{^{''}} + 6 y^{^{'}} + 9 y = 18 t^{2} + 24 t + 4