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Popular >

(x+y)y^'=1

Solution

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

Solution

y = - W (\frac{c _{1}}{e ^{x + 1}}) - x - 1

Solution steps

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

Solve exact ODE:

y = - W (\frac{c _{1}}{e ^{x + 1}}) - x - 1

y = - W (\frac{c _{1}}{e ^{x + 1}}) - x - 1

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\int_{1}^{7} π (x - 1)^{2} d x

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Frequently Asked Questions (FAQ)

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