t^2y^'+y^'=ty

Lösung

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

y = c_{1} t^{2} + 1

Schritte zur Lösung

t^{2} y^{'} (t) + y^{'} (t) = t y

Solve separable ODE:

y = c_{1} t^{2} + 1

y = c_{1} t^{2} + 1

d er i v a t i v e cos (sin (cot (π x)))

t an g e n t 2 x + \frac{3}{x}

\int_{0}^{4} ∣ x - 3 ∣ d x

d er i v a t i v e f (x) = \frac{4 x ^{2} + 4 x + 6}{x}

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