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Popular >

y^'=2(t+1)y,y(0)=1

Solution

y^{^{'}} = 2 (t + 1) y, y (0) = 1

Solution

y = e^{t^{2} + 2 t}

Solution steps

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

Solve separable ODE:

y = e^{t^{2} + 2 t}

y = e^{t^{2} + 2 t}

Popular Examples

y^{''}+2*y^'+y=e^{-t}

y^{^{''}} + 2 \cdot y^{^{'}} + y = e^{- t}

sum from n=0 to infinity of 1/(2^nn!)

n = 0 \sum \infty \frac{1}{2 ^{n} n !}

derivative of f(x)=((x^3+1))/((x^4+1))derivative of

f (x) = \frac{( x ^{3} + 1 )}{( x ^{4} + 1 )}

limit as x approaches-1 of (x+1)/(x^2-1)

x \to - 1 lim (\frac{x + 1}{x ^{2} - 1})

integral of 2sqrt(4-x^2)

\int 2 4 - x^{2} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=2(t+1)y,y(0)=1 ?
The solution for y^'=2(t+1)y,y(0)=1 is y=e^{t^2+2t}