What is the solution for y(x+7)+y^'=0,y(-14)=1 ?

The solution for y(x+7)+y^'=0,y(-14)=1 is y=e^{-(x^2)/2-7x}

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y(x+7)+y^'=0,y(-14)=1

y (x + 7) + y^{^{'}} = 0, y (- 14) = 1

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (7 cos (7 x))

\int_{0}^{2 π} x sin (x) d x

\int - 7 x d x

x \to - 2 + lim (\frac{x + 2}{x ^{3} + 8})

\frac{1}{x - 9}

What is the solution for y(x+7)+y^'=0,y(-14)=1 ?
The solution for y(x+7)+y^'=0,y(-14)=1 is y=e^{-(x^2)/2-7x}