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Popular Calculus >

7+e^{-5x}y^'=0

Solution

7 + e^{- 5 x} y^{^{'}} = 0

Solution

y = - \frac{7}{5} e^{5 x} + c_{1}

Solution steps

7 + e^{- 5 x} y^{'} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = - 7 e^{5 x}

If

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

then

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

y = \int - 7 e^{5 x} d x

\int - 7 e^{5 x} d x = - \frac{7}{5} e^{5 x} + c_{1}

y = - \frac{7}{5} e^{5 x} + c_{1}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for 7+e^{-5x}y^'=0 ?
The solution for 7+e^{-5x}y^'=0 is y=-7/5 e^{5x}+c_{1}