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Popular >

(x^2-1)y^'=y

Solution

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

Solution

y = \frac{c _{1} x - 1}{x + 1}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1} x - 1}{x + 1}

y = \frac{c _{1} x - 1}{x + 1}

Popular Examples

(dy)/(dx)y=sqrt(u),u=x^2+10

\frac{d y}{d x} y = u, u = x^{2} + 10

y^{^{'}} + 5 y + 4 y = 0

(dy)/(dx)+y/(50)=4,y(0)=30

\frac{d y}{d x} + \frac{y}{50} = 4, y (0) = 30

(dV)/(dt)=80-8/100 V

\frac{d V}{d t} = 80 - \frac{8}{100} V

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

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

Frequently Asked Questions (FAQ)

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