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Popular >

y^'=((1+y))/(x^2+x)

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

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y^{''}-2y+y=t+1,y(0)=0,y^'(0)=0

y^{^{''}} - 2 y + y = t + 1, y (0) = 0, y^{^{'}} (0) = 0

Frequently Asked Questions (FAQ)

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