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Popular Calculus >

y^'=10^{x+y}

Solution

y^{^{'}} = 1 0^{x + y}

Solution

y = - \frac{ln ( - 1 0 ^{x} - c _{1} )}{ln ( 10 )}

Solution steps

y^{'} (x) = 1 0^{x + y}

Solve separable ODE:

y = - \frac{ln ( - 1 0 ^{x} - c _{1} )}{ln ( 10 )}

y = - \frac{ln ( - 1 0 ^{x} - c _{1} )}{ln ( 10 )}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=10^{x+y} ?
The solution for y^'=10^{x+y} is y=-(ln(-10^x-c_{1}))/(ln(10))