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Popular >

y^'=-y,y(0)=1

Solution

y^{^{'}} = - y, y (0) = 1

Solution

y = e^{- t}

Solution steps

y^{'} (t) = - y

Solve separable ODE:

y = e^{- t}

y = e^{- t}

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g (x) = \frac{2 x + 2}{x ^{2} + 2}

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n = 0 \sum \infty n

limit as (x,y) approaches (-6,5) of (x^2cos(piy))/(e^{x+y)}

(x, y) \to (- 6, 5) lim (\frac{x ^{2} cos ( π y )}{e ^{x + y}})

(d^2)/(dx^2)(cos(pix))

\frac{d ^{2}}{d x ^{2}} (cos (π x))

integral of 8/(x^2+9)

\int \frac{8}{x ^{2} + 9} d x

Frequently Asked Questions (FAQ)

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