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Beliebt Rechnen >

y^'+y=y(te^{t^2}+1)t,y(0)=1

Lösung

y^{^{'}} + y = y (t e^{t^{2}} + 1) t, y (0) = 1

Lösung

y = e^{\frac{t ^{2}}{2} + \frac{e ^{t^{2}} t}{2} - t - \frac{π}{4} erfi (t)}

Schritte zur Lösung

y^{'} (t) + y = y (t e^{t^{2}} + 1) t

Solve separable ODE:

y = e^{\frac{t ^{2}}{2} + \frac{e ^{t^{2}} t}{2} - t - \frac{π}{4} erfi (t)}

y = e^{\frac{t ^{2}}{2} + \frac{e ^{t^{2}} t}{2} - t - \frac{π}{4} erfi (t)}

Beliebte Beispiele

(dy)/(dx)y=(x^{x-3})

\frac{d y}{d x} y = (x^{x - 3})

((x^3+1)/(x^2))(dy)/(dx)= 1/y ,y(0)=-1

(\frac{x ^{3} + 1}{x ^{2}}) \frac{d y}{d x} = \frac{1}{y}, y (0) = - 1

x^{^{'}} = x^{7}

(dv}{dt}=\frac{0.1)/v*(0.25)/t

\frac{d v}{d t} = \frac{0.1}{v} \cdot \frac{0.25}{t}

y(dy)/(dx)=3,x(y^2+6)^8=3

y \frac{d y}{d x} = 3, x (y^{2} + 6)^{8} = 3