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Popular >

x^4-x+y-xy^'=0

Solution

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

Solution

y = \frac{x ^{4}}{3} - x ln (x) + c_{1} x

Solution steps

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

Solve linear ODE:

y = \frac{x ^{4}}{3} - x ln (x) + c_{1} x

y = \frac{x ^{4}}{3} - x ln (x) + c_{1} x

Popular Examples

(\partial)/(\partial x)(e^{4x+5y})

\frac{\partial}{\partial x} (e^{4 x + 5 y})

derivative of 1+3cos^2(x)

\frac{d}{d x} (1 + 3 cos^{2} (x))

integral of (2x-x^2-1)/(x^3+x)

\int \frac{2 x - x ^{2} - 1}{x ^{3} + x} d x

derivative of 7-5x

\frac{d}{d x} (7 - 5 x)

tangent of y=arcsin(x/2),\at x=1 tangent of

y = arcsin (\frac{x}{2}), at x = 1

Frequently Asked Questions (FAQ)

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