What is the solution for y^'=(y+5)(x+2),y(0)=1 ?

The solution for y^'=(y+5)(x+2),y(0)=1 is y=6e^{(x^2)/2+2x}-5

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y^'=(y+5)(x+2),y(0)=1

y^{^{'}} = (y + 5) (x + 2), y (0) = 1

y = 6 e^{\frac{x ^{2}}{2} + 2 x} - 5

Solution steps

y^{'} (x) = (y + 5) (x + 2)

Solve separable ODE:

y = 6 e^{\frac{x ^{2}}{2} + 2 x} - 5

y = 6 e^{\frac{x ^{2}}{2} + 2 x} - 5

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What is the solution for y^'=(y+5)(x+2),y(0)=1 ?
The solution for y^'=(y+5)(x+2),y(0)=1 is y=6e^{(x^2)/2+2x}-5