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Popular >

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

Solution

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

Solution

y = t^{2}

Solution steps

y^{'} (t) = 2 t

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 2 t d t

\int 2 t d t = t^{2} + c_{1}

y = t^{2} + c_{1}

Apply initial conditions:

y = t^{2}

y = t^{2}

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\int tan^{6} (a x) d x

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Frequently Asked Questions (FAQ)

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