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Popular >

y^'=(t^3)/y ,y(0)=4

Solution

y^{^{'}} = \frac{t ^{3}}{y}, y (0) = 4

Solution

y = \frac{t ^{4}}{2} + 16

Solution steps

y^{'} (t) = \frac{t ^{3}}{y}

Solve separable ODE:

y = \frac{t ^{4}}{2} + 16

y = \frac{t ^{4}}{2} + 16

Popular Examples

(dy)/(dx)=x^2+x^2y

\frac{d y}{d x} = x^{2} + x^{2} y

(dv)/(dt)=sqrt(9.8v),v(0)=5.3

\frac{d v}{d t} = 9.8 v, v (0) = 5.3

(dx)/(dt)=100-x/(100),x=4000,t=0

\frac{d x}{d t} = 100 - \frac{x}{100}, x = 4000, t = 0

d v = e^{3 x} d x

(6+x)y^'=(x+7)(y-6)

(6 + x) y^{^{'}} = (x + 7) (y - 6)

Frequently Asked Questions (FAQ)

What is the solution for y^'=(t^3)/y ,y(0)=4 ?
The solution for y^'=(t^3)/y ,y(0)=4 is y=sqrt((t^4)/2+16)