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Popular >

e^ty^'=(1+y^2)t

Solution

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

Solution

y = tan (\frac{- t - 1}{e ^{t}} + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (\frac{- t - 1}{e ^{t}} + c_{1})

y = tan (\frac{- t - 1}{e ^{t}} + c_{1})

Popular Examples

x^2y^'=y^2,y(1)=-1

x^{2} y^{^{'}} = y^{2}, y (1) = - 1

xydy=(1+y^2)(1-x)dx

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

y^'= x/y ,y(0)=-3

y^{^{'}} = \frac{x}{y}, y (0) = - 3

d/(dt)(\H(t))=4-0.5\H(t),\H(0)=0

\frac{d}{d t} (H (t)) = 4 - 0.5 H (t), H (0) = 0

(dy)/(dx)=(10+16x)/(xy^2),y=4,x=1

\frac{d y}{d x} = \frac{10 + 16 x}{x y ^{2}}, y = 4, x = 1

Frequently Asked Questions (FAQ)

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