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Popular >

y^'=y^2t^3

Solution

y^{^{'}} = y^{2} t^{3}

Solution

y = - \frac{4}{t ^{4} + c _{1}}

Solution steps

y^{'} (t) = y^{2} t^{3}

Solve separable ODE:

y = - \frac{4}{t ^{4} + c _{1}}

y = - \frac{4}{t ^{4} + c _{1}}

Popular Examples

\frac{d V}{d t} = 0.4 V

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

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

(3y^2x^2-3x^2)dx+(5x^3y)dy=0

(3 y^{2} x^{2} - 3 x^{2}) d x + (5 x^{3} y) d y = 0

(dy)/(dx)=ln(x^y),y(1)=1

\frac{d y}{d x} = ln (x^{y}), y (1) = 1

(y^2+y)dx-xdy=0,y(1)=1

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^2t^3 ?
The solution for y^'=y^2t^3 is y=-4/(t^4+c_{1)}