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Popular >

(x^2-x)y^'+(3-2x)y=0

Solution

(x^{2} - x) y^{^{'}} + (3 - 2 x) y = 0

Solution

y = \frac{c _{1} x ^{3}}{x - 1}

Solution steps

(x^{2} - x) y^{'} (x) + (3 - 2 x) y = 0

Solve separable ODE:

y = \frac{c _{1} x ^{3}}{x - 1}

y = \frac{c _{1} x ^{3}}{x - 1}

Graph

Popular Examples

y^'sec(t)=(-y^9)/8

y^{^{'}} sec (t) = \frac{- y ^{9}}{8}

(1+e^x)y-(x+e^x)y^'=0

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

y^'=(x^2)/(y^2),y(2)=4

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

(dy)/(dx)=(2x+2xy^2)/(y+2x^2y)

\frac{d y}{d x} = \frac{2 x + 2 x y ^{2}}{y + 2 x ^{2} y}

y^'=xy-7x,y(0)=8

y^{^{'}} = x y - 7 x, y (0) = 8

Frequently Asked Questions (FAQ)

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