At any point `P(x,y)` of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact `P` to the point `(-4,-3)` Find the equation of the curve given that it passes through the point `(-2, 1)`

Explanation

Let A ,(x,y) by any point on the curve y= f (x)
Then slope of the tangent to the curve at the point A is ` (dy)/dx`
According to the given condition.
` dy/dx +y +2 x`
` dy /dx -y =2x `
This is the linear differential equation of the form .
` (dy)/(dx) + Py =Q " where " P = -1 and Q = 2x`
` I.F .= e^(int p dx) =e^(int -1dx) =e^(-x)`
The solution of (1) is given by
`y.(I.F)= int Q. (I.F)dx+c`
`therefore e^(-x).y =2int x.e^(-x) dx +c`
` e^(-x).y =2[ x.int e^(-x)dx -int { d/(dx) int e^(-x)dx} dx ]+c`
` e^(-x) .y=2[x.(e^(x))/(-1) - int 1. (e^(-x))/(-1)dx]+c`
` e^(-x).y = -2xe^(-x) .y2 [ x.(6^(-x))/(-1) - int 1. (e^(-x))/(-1) dx]+c`
`e^(-x).y = -2xe^(-x) + 2int e^(-x) dx+c`
` e^(x).y = -2xe^(-x) +2int e^(-x) dx+c`
`e^(-x)y= -2xe^(-x) + 2.(e^(-x))/(-1) +c`
` y = -2x -2 +ce^(x)`
` y+2(x+1) =ce^(x)`
this is the general equation of the curve.
But the required curve is passing through the origin (0,0). ltbgt put x =0 and y= 0 in (2), we get ,
0 + 2(0+1)=c , `therefore c=2`
From (2). the equation of the required curve is
`y+ 2(x+1) =2e^(x)`