y^'=3x+y,y(0)=6

Lösung

y^{^{'}} = 3 x + y, y (0) = 6

y = - 3 (x + 1) + 9 e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = - 3 (x + 1) + 9 e^{x}

y = - 3 (x + 1) + 9 e^{x}

p a r i t y y = ln (e^{- x} + x e^{- x})

\frac{\partial}{\partial x} (- \frac{c x}{x ^{2} + y ^{2}})

\int_{1}^{3} \frac{1}{4 + ( x - 1 ) ^{2}} d x

\int 24 x y^{2} + 3 y e^{x y} d x

\frac{d y}{d x} = e^{(6 x - 3 y)}, y (0) = 3