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Popular >

y^'(y-x)=1

Solution

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

Solution

y = x + c_{1}

Solution steps

y^{'} (x) = 1

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 1 d x

\int 1 d x = x + c_{1}

y = x + c_{1}

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Frequently Asked Questions (FAQ)

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