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Popular >

y^'=(y-1)/x

Solution

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

Solution

y = c_{1} x + 1

Solution steps

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

Solve separable ODE:

y = c_{1} x + 1

y = c_{1} x + 1

Graph

Popular Examples

(dy)/(dx)=-2(y-40)

\frac{d y}{d x} = - 2 (y - 40)

3+2q+(4+2p)(dq)/(dp)=0

3 + 2 q + (4 + 2 p) \frac{d q}{d p} = 0

sin(3x)dx+2ycos(33x)dy=0

sin (3 x) d x + 2 y cos (33 x) d y = 0

cos(x)y^'= y/(ln(y)),y(0)=1

cos (x) y^{^{'}} = \frac{y}{ln ( y )}, y (0) = 1

(dy)/(dx)=-(4*+3y+15)/(2*+y+7)

\frac{d y}{d x} = - \frac{4 \cdot + 3 y + 15}{2 \cdot + y + 7}

Frequently Asked Questions (FAQ)

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