(x+4)^2y^'+3(x+4)y=4

Lösung

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

y = \frac{2 ( x ^{2} + 8 x )}{x ^{3} + 12 x ^{2} + 48 x + 64} + \frac{c _{1}}{x ^{3} + 12 x ^{2} + 48 x + 64}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{2 ( x ^{2} + 8 x )}{x ^{3} + 12 x ^{2} + 48 x + 64} + \frac{c _{1}}{x ^{3} + 12 x ^{2} + 48 x + 64}

y = \frac{2 ( x ^{2} + 8 x )}{x ^{3} + 12 x ^{2} + 48 x + 64} + \frac{c _{1}}{x ^{3} + 12 x ^{2} + 48 x + 64}

\int \frac{50}{( x - 1 ) ( x ^{2} + 49 )} d x

d er i v a t i v e ln (e^{x^{2}})

\int \frac{x ^{2}}{x ^{2} - 4 x + 13} d x

\int \frac{e ^{4 x}}{2} d x

x \to 0 lim (x \cdot ln (x))