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Popular >

y^'+t^2=0

Solution

y^{^{'}} + t^{2} = 0

Solution

y = - \frac{t ^{3}}{3} + c_{1}

Solution steps

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

Isolate

y^{'} (t) : y^{'} (t) = - t^{2}

If

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

then

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

y = \int - t^{2} d t

\int - t^{2} d t = - \frac{t ^{3}}{3} + c_{1}

y = - \frac{t ^{3}}{3} + c_{1}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+t^2=0 ?
The solution for y^'+t^2=0 is y=-(t^3)/3+c_{1}