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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

y^'=(xsqrt(x^2-1))/y ,y(-1)=1

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

1/y (dy)/(dx)=(1-2x)

\frac{1}{y} \frac{d y}{d x} = (1 - 2 x)

(dy)/(dx)=e^y,y=0,x=1

\frac{d y}{d x} = e^{y}, y = 0, x = 1

x^{2} y^{^{'}} = (x + 7) y

(dx)/(dy)=(x^3+2x^2+1)/(y^2)

\frac{d x}{d y} = \frac{x ^{3} + 2 x ^{2} + 1}{y ^{2}}

Frequently Asked Questions (FAQ)

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