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Popular >

y^'=ty^2,y(0)=2

Solution

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

Solution

y = - \frac{2}{t ^{2} - 1}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{t ^{2} - 1}

y = - \frac{2}{t ^{2} - 1}

Popular Examples

(dy)/(dt)=0.03(291-y)

\frac{d y}{d t} = 0.03 (291 - y)

(dy)/(dx)=(cos(5x))/(4y^3)

\frac{d y}{d x} = \frac{cos ( 5 x )}{4 y ^{3}}

y^{^{'}} + 2 y = 4 x y

(dx}{dy}=\frac{(x+2)^2)/x

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

(1+x^2)(dy)/(dx)-xy=0,y(2)=5

(1 + x^{2}) \frac{d y}{d x} - x y = 0, y (2) = 5

Frequently Asked Questions (FAQ)

What is the solution for y^'=ty^2,y(0)=2 ?
The solution for y^'=ty^2,y(0)=2 is y=-2/(t^2-1)