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Popular >

(1+x)y^'-y=0

Solution

(1 + x) y^{^{'}} - y = 0

Solution

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

Solution steps

(1 + x) y^{'} (x) - y = 0

Solve separable ODE:

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

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

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Frequently Asked Questions (FAQ)

What is the solution for (1+x)y^'-y=0 ?
The solution for (1+x)y^'-y=0 is y=e^{c_{1}}+e^{c_{1}}x