What is the solution for y(x+9)+y^'=0,y(-18)=1 ?

The solution for y(x+9)+y^'=0,y(-18)=1 is y=e^{-(x^2)/2-9x}

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y(x+9)+y^'=0,y(-18)=1

y (x + 9) + y^{^{'}} = 0, y (- 18) = 1

y = e^{- \frac{x ^{2}}{2} - 9 x}

Solution steps

y (x + 9) + y^{'} (x) = 0

Solve separable ODE:

y = e^{- \frac{x ^{2}}{2} - 9 x}

y = e^{- \frac{x ^{2}}{2} - 9 x}

\frac{\partial}{\partial x} (x^{2} + y^{2} + z)

y = x (x^{4} - x^{3})

y = \frac{36 x}{x ^{2} + 36}, (- 3, - \frac{12}{5})

\frac{d}{d x} (21 x + x^{2})

\frac{d}{d x} (e^{2 x})

What is the solution for y(x+9)+y^'=0,y(-18)=1 ?
The solution for y(x+9)+y^'=0,y(-18)=1 is y=e^{-(x^2)/2-9x}