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Popular >

y(x+7)+y^'=0

Solution

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

Solution

y = e^{- \frac{x ^{2}}{2} - 7 x + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{- \frac{x ^{2}}{2} - 7 x + c_{1}}

y = e^{- \frac{x ^{2}}{2} - 7 x + c_{1}}

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\frac{d}{d x} (tan (5 x + 2))

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tangent of x^4+xy^2=5,(1,2)tangent of

x^{4} + x y^{2} = 5, (1, 2)

derivative of 4x-2/(x^3)

\frac{d}{d x} (4 x - \frac{2}{x ^{3}})

Frequently Asked Questions (FAQ)

What is the solution for y(x+7)+y^'=0 ?
The solution for y(x+7)+y^'=0 is y=e^{-(x^2)/2-7x+c_{1}}