Differential Equations · Question 4

Let y(x) be the solution of the differential equation x2 dydx+xy=x2+y2, x> 1e, satisfying y(1)=0. Then the value of 2 (y(e))2y(e2) is .

Answer

0.7 to 0.8

Solution

Homogeneous: put y=vx: x2(v+xv')+x2v=x2+x2v2 x dvdx=(v-1)2.

dv(v-1)2= dxx 11-v= e x+C; y(1)=0 v=0 at x=1 C=1.

v= e x1+ e x y= x e x1+ e x.

y(e)= e2, y(e2)= 2e23.

2 e2/42e2/3= 34=0.75

Answer: 0.75