Question

The slope of the tangent to the curve at any point is equal to y+ 2x. Find the equation of the curve passing through the origin.

Answer 1

 

 Since dy/dx represents the slope of a tangent at any point (x, y) on a given curve, according to given condition 

`dy/dx=y+2x`

`dy/dx-y=2x` which is in the form of linear differential equation `dy/dx+Py=Q`

where P=-1 ,Q=2x

`I.f=e^(intPdx)=e^(int-1dx)=e^(-x)`

The solution is given by

`y.e^(-x)=intQ(IF)dx+C'`

`y.e^(-x)=2intx.e^(-x) dx+c`

Consider `int x.e^(-x)dx` ,Integrating by part

`=x e^(-x)/-1 -int[d/dx x inte^(-x)dx]dx`

`=-xe^(-x)-inte^(-x)/-1dx`

`=-xe^(-x)-e^(-x)`

`ye^(-x)=2(-xe^(-x)-e^(-x))+c`

`y=-2x-2+ce^x`

`2x + y + 2 = ce^x` is the general solution. Since the curve is passing through origin, x = 0, y = 0

`0 + 0+ 2 =c.e^0 => c = 2`

`2x +y + 2 =2e^x`

 is the equation of the required curve.