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Popular >

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

Solution

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

Solution

y = ln (1 + x) (1 + x) + 1 + c_{1} (1 + x)

Solution steps

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

Solve linear ODE:

y = ln (1 + x) (1 + x) + 1 + c_{1} (1 + x)

y = ln (1 + x) (1 + x) + 1 + c_{1} (1 + x)

Graph

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Frequently Asked Questions (FAQ)

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