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Popular >

y^'=(1+x)/y ,y(0)=5

Solution

y^{^{'}} = \frac{1 + x}{y}, y (0) = 5

Solution

y = 2 x + x^{2} + 25

Solution steps

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

Solve separable ODE:

y = 2 x + x^{2} + 25

y = 2 x + x^{2} + 25

Graph

Popular Examples

y^{^{'}} sin (y) = 1

(dx)/(dt)=x^2+1/16 ,x(0)=6

\frac{d x}{d t} = x^{2} + \frac{1}{16}, x (0) = 6

(dv)/(dx)=9+16v^2

\frac{d v}{d x} = 9 + 16 v^{2}

y^'+(2xy^2+2y)/(2x^2y+2x)=0

y^{^{'}} + \frac{2 x y ^{2} + 2 y}{2 x ^{2} y + 2 x} = 0

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=(1+x)/y ,y(0)=5 ?
The solution for y^'=(1+x)/y ,y(0)=5 is y=sqrt(2x+x^2+25)