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Popular >

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

Solution

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

Solution

y = - ln (- \frac{1}{7} e^{7 x} - c_{1})

Solution steps

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

Solve separable ODE:

y = - ln (- \frac{1}{7} e^{7 x} - c_{1})

y = - ln (- \frac{1}{7} e^{7 x} - c_{1})

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Frequently Asked Questions (FAQ)

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