Toronto Math Forum

solve the given initial value problem:
y''+4y'+4y=1, y(-1)=2, y'(-1)=1

solution:
r^2+4r+4=0
(r+2)^2=0
r1=r2=-2
Repeated roots
Homo solution is
y(t)=(c1)e^(-2t)+(c2)e^(-2t) (t)
y'(t)=-2(c1) e^(-2t)+(c2) e^(-2t)-2t(c2) e^(-2t)
Plug in initial value y(-1)=2 and y’(-1)=1
c2=5/(e^2)
c1=7/(e^2)

y(t)=(7/e^2)e^(-2t)+(5/e^2)e^(-2t) (t)

(sin⁡y/y-2e^(-x)  sin⁡x )+((cos⁡y+2e^(-x)  cos⁡x)/y)y'=0, u(x,y)=ye^x

Answer:
(sin⁡y/y-2e^(-x)  sin⁡x )+((cos⁡y+2e^(-x)  cos⁡x)/y)y'=0
M=sin⁡y/y-2e^(-x)  sin⁡x
dM/dY=〖-y〗^(-2)  sin⁡y+1/y  cos⁡y
N=(cos⁡y+2e^(-x)  cos⁡x)/y
dN/dX=1/y (-2e^(-x)  cos⁡x+2e^(-x) (-sin⁡x ))

Since dM/dY does not equal dN/dX
The equation is not exact.
Multiply μ=ye^x to both side of the equation
Get:

sin⁡〖(y) e^x 〗-2y sin⁡x+(e^x  cos⁡y+2 cos⁡(x))y^'=0
M=sin⁡〖(y) e^x 〗-2y sin⁡x
dM/dY=e^x  cos⁡y-2 sin⁡x
N=e^x  cos⁡y+2 cos⁡x
dN/dX=e^x  cos⁡y-2 sin⁡x

Now dM/dY=dN/dX
The equation is exact.

There exists ϕ(x,y) s.t. dϕ/dx=M, dϕ/dy=N

ϕ=∫M dx=∫〖sin⁡〖(y)〗 e^x 〗-2y sin⁡(x) dx=sin⁡(y) e^x +2y cos⁡(x)+h(y)
dϕ/dy=e^x  cos⁡y+2 cos⁡x+h'(y)
Also, dϕ/dy=N=e^x  cos⁡y+2 cos⁡x
So h’(y)=0

ϕ(x,y)=sin⁡(y) e^x +2y cos⁡(x)=C