Let y_1 and y_2 be solution of a second order homogeneous linear differential equation y''+p(x)y'+q(x)=0, in R. Suppose that y_1(x)+y_2(x)=e^{-x}, W[y_1(x),y_2(x)]=e^x, where W[y_1,y_2] is the Wroian of y_1 and y_2. Find p(x), q(x) and the general form of y_1 and y_2.

Rivka Thorpe 2020-11-09 Answered
Let y1 and y2 be solution of a second order homogeneous linear differential equation y+p(x)y+q(x)=0, in R. Suppose that y1(x)+y2(x)=ex,
W[y1(x),y2(x)]=ex, where W[y1,y2] is the Wro
ian of y1 and y2.
Find p(x), q(x) and the general form of y1 and y2.
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Expert Answer

delilnaT
Answered 2020-11-10 Author has 94 answers
y+p(x)y+q(x)=0 (*)
y1(x)+y2(x)=ex
W(y1,y2)=ex
Since y1, and y2 are solution to (*)
So their linear combinations are also a solution to (*)
Hence y1+y2=ex is also a solution to (*)
Now,
(y1+y2)=ex
(y1+y2)""=ex
So, from (*) we have
(y1+y2)+p(x)(y1+y2)+q(x)(y1+y2)=0
exexp(x)+exq(x)=0
or ex[1p(x)+q(x)]=0
As ex0  x
Hence, 1p(x)+q(x)=0 (I)
Now, W[y1,y2]=[(y1,y2),(y1,y2)]=y1y2y1y2=ex (II)
As y1+y2=ex
y2=exy1
and y1+y2=ex
so, y2=exy1
Putting the value of y2 and y2 in (II) we got,
y1(exy1)y1(exy1)=ex
exy1y1y1exy1+y1y1=ex
ex[y1+y1]=ex
y1+y1=e2x
y1=e2xy1
y1=2e2xy1
y1=2e2xy1=2e2x(e2xy1)
=e2x+y1
y1=e2x+y1
Putting the value of y1 and y1 in (*) we get,
(e2x+y1)+p(x)(e2xy1)+q(x)y1=0
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