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Popular >

2x^4y^'=k

Solution

2 x^{4} y^{^{'}} = k

Solution

y = - \frac{k}{6 x ^{3}} + c_{1}

Solution steps

2 x^{4} y^{'} (x) = k

Isolate

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

If

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

then

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

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

\int \frac{k}{2 x ^{4}} d x = - \frac{k}{6 x ^{3}} + c_{1}

y = - \frac{k}{6 x ^{3}} + c_{1}

Popular Examples

integral of (2x^3+6x)^2

\int (2 x^{3} + 6 x)^{2} d x

(\partial)/(\partial x)(5ln(sqrt(xy)))

\frac{\partial}{\partial x} (5 ln (x y))

x^{''}+20x^'+500x=12

x^{^{''}} + 20 x^{^{'}} + 500 x = 12

derivative of y=x^2(1-2x)derivative of

y = x^{2} (1 - 2 x)

integral from 0 to 4 of sqrt(x)-x+2

\int_{0}^{4} x - x + 2 d x

Frequently Asked Questions (FAQ)

What is the solution for 2x^4y^'=k ?
The solution for 2x^4y^'=k is y=-k/(6x^3)+c_{1}