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Popular >

15x^2-4y^'=0

Solution

15 x^{2} - 4 y^{^{'}} = 0

Solution

y = \frac{5 x ^{3}}{4} + c_{1}

Solution steps

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

Isolate

y^{'} (x) : y^{'} (x) = \frac{15 x ^{2}}{4}

If

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

then

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

y = \int \frac{15 x ^{2}}{4} d x

\int \frac{15 x ^{2}}{4} d x = \frac{5 x ^{3}}{4} + c_{1}

y = \frac{5 x ^{3}}{4} + c_{1}

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Frequently Asked Questions (FAQ)

What is the solution for 15x^2-4y^'=0 ?
The solution for 15x^2-4y^'=0 is y=(5x^3)/4+c_{1}