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Popular >

y^'=y^2+2

Solution

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

Solution

y = tan (2 (t + c_{1})) 2

Solution steps

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

Solve separable ODE:

y = tan (2 (t + c_{1})) 2

y = tan (2 (t + c_{1})) 2

Popular Examples

(dy)/(dx)=-4/4 x^8y^4

\frac{d y}{d x} = - \frac{4}{4} x^{8} y^{4}

(dy)/(dt)+2(t+1)y^2=0,y(0)=-1/8

\frac{d y}{d t} + 2 (t + 1) y^{2} = 0, y (0) = - \frac{1}{8}

y^'=(ln(x))/(xy),y(1)=2

y^{^{'}} = \frac{ln ( x )}{x y}, y (1) = 2

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

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

(dy)/(dt)=e^{-y}(2t-4),y(2)=0

\frac{d y}{d t} = e^{- y} (2 t - 4), y (2) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^2+2 ?
The solution for y^'=y^2+2 is y=tan(sqrt(2)(t+c_{1}))sqrt(2)