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Popular >

-4y^'+12x^2=0

Solution

- 4 y^{^{'}} + 12 x^{2} = 0

Solution

y = x^{3} + c_{1}

Solution steps

- 4 y^{'} (x) + 12 x^{2} = 0

Isolate

y^{'} (x) : y^{'} (x) = 3 x^{2}

If

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

then

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

y = \int 3 x^{2} d x

\int 3 x^{2} d x = x^{3} + c_{1}

y = x^{3} + c_{1}

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Frequently Asked Questions (FAQ)

What is the solution for -4y^'+12x^2=0 ?
The solution for -4y^'+12x^2=0 is y=x^3+c_{1}