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Popular >

y^'-y/x-1= 1/x+y

Solution

y^{^{'}} - \frac{y}{x} - 1 = \frac{1}{x} + y

Solution

y = e^{x + c_{1}} x - 1

Solution steps

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

Solve separable ODE:

y = e^{x + c_{1}} x - 1

y = e^{x + c_{1}} x - 1

Popular Examples

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

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

(dy)/(dt)+2y=0,y(0)=3

\frac{d y}{d t} + 2 y = 0, y (0) = 3

(dy)/(dx)+6x^2y=5x^2,y(0)=4

\frac{d y}{d x} + 6 x^{2} y = 5 x^{2}, y (0) = 4

y^{^{'}} = x y - 2 x + y - 2

(dy)/(dx)=(x+1)/(e^y),y(0)=2

\frac{d y}{d x} = \frac{x + 1}{e ^{y}}, y (0) = 2

Frequently Asked Questions (FAQ)

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