What is the solution for (y+4)y^'=2xy+4y,y(2)=1 ?

The solution for (y+4)y^'=2xy+4y,y(2)=1 is y+4ln(y)=x^2+4x-11

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(y+4)y^'=2xy+4y,y(2)=1

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

y + 4 ln (y) = x^{2} + 4 x - 11

Solution steps

(y + 4) y^{'} (x) = 2 x y + 4 y

Solve separable ODE:

y + 4 ln (y) = x^{2} + 4 x - 11

y + 4 ln (y) = x^{2} + 4 x - 11

\int (1 + 30 w) w d w

\int 1 - x^{2} d x

f (x) = 6 x + 8

\frac{d}{d x} (arctan (7 x^{2} - 1))

What is the solution for (y+4)y^'=2xy+4y,y(2)=1 ?
The solution for (y+4)y^'=2xy+4y,y(2)=1 is y+4ln(y)=x^2+4x-11