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Popular >

y^'=x(1+y)

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

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y = \frac{9000}{x ^{2}}

(\partial)/(\partial x)(4xyz+ln(2xyz))

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Frequently Asked Questions (FAQ)

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