Solve the following IVP for the exact equation and find the interval of validity for the solution. 2xy−9x2+(2y+x2+1) dydx=0 y(0)=−2

Name: Solve the following IVP for the exact equation and find the interval of validity for the solution. 2xy−9x2+(2y+x2+1) dydx=0 y(0)=−2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the following IVP for the exact equation and find the interval of validity for the solution. 2xy−9x2+(2y+x2+1) dydx=0 y(0)=−2

Question

Solve the following IVP for the exact equation and find the interval of validity for the solution.  2xy−9x2+(2y+x2+1) dydx=0 y(0)=−2

Barbara Meeker · Accepted Answer

For any differential equation that is given to be exact, we must have a function f(x,y) such that:  d[f(x,y)]=∂f∂xdx+∂f∂ydy=M dx+N dy  ∴M(x,y)=∂f∂x and N(x,y)=∂f∂y  i.e.f(x,y)=∫M(x,y)dx+ϕ1(y) and f(x,y)=∫N(x,y)dy+ϕ2(x)  The given differential equation can be written as:  2xy−9x2+(2y+x2+1)dydx=0  (2xy−9x2)dx+(2y+x2+1)dy=0 (i)  Comparing to the standard form of M(x,y)dx+N(x.y)dy=0 we get:  M(x,y)=2xy−9x2  ∴f(x,y)=∫(2xy−9x2)dx+ϕ1(y)  =2y(x22)−9(x33)+ϕ1(y)  =x2y−3x3+ϕ1(y)...(2)  N(x,y)=2y+x2+1  ∴f(x,y)=∫(2y+x2+1)dy+ϕ2(x)  =2(y22)+x2(y)+(y)+ϕ2(x)

reinosodairyshm · Accepted Answer

[y(2x)dx+x2(dy)]+(2y+1)dy−9x2dx=0  d(x2y)+d(y2+y)−d(3x3)=0  ⇒d(x2y+y2+y−3x2)=0  ⇒x2y+y2+y−3x2=c  y(0)=−3⇒0+4−3+0=c=6⇒x2y+y2+y−3x2=6

karton · Accepted Answer

(2xy−9x2)dx+(2y+x2+1)dy=0My=2x=Nx=2x∫M dx+g(y)H(x,y)=∫2xy−9x2dx+g(y)=∫2y+x2+1 dy=x2y−3x3+g(y)=y2+x2y+y Since x2y appears on both sides, cancel it on the yside, which gives g(y)=y2+y. Add this to the leftside and the general solution is: H(x,y)=x2y−3x2+y2+y