Let u be a solution of the differential

Zack Mora

Zack Mora

Answered question

2022-03-21

Let u be a solution of the differential equation y+xy=0 and ϕ=uψ be a solution of the differential equation y+2xy+(x2+2)y=0 such that ϕ(0)=1,ϕ(0)=0. Then ϕ(x) is equal to
A) (cos2(x))ex22
B) (cos(x))ex22
C) (1+x2)ex22
D) (cos(x))ex2

Answer & Explanation

Chad Robinson

Chad Robinson

Beginner2022-03-22Added 6 answers

Step 1
It's almost tempting to use reduction of order here but unfortunately that doesn't work here as u is not a solution of y +2xy+(x2+2)y=0. So now the natural way is to use the given fact that ϕ is a solution of the blue colored ODE while noting that the blue colored ODE is a homogeneous one so there is some 'hope' of cancellation of some terms. You are almost there.
Step 2
Consider u=ex22 (just taking c=1 in u that you obtained and this can be done due to homogeneity of the blue colored ODE). Then note that u=xex22 and u =(x21)ex22. Substitute ϕ=uψ in y +2xy+(x2+2)y=0 to get
  (uψ+uψ+2uψ)+2x(uψ+uψ)+(x2+2)uψ=ex22(ψ+ψ) It follows that ex22(ψ+ψ)=0 which gives ψ+ψ=0  ψ(x)=acosx+bsinx, where a and b are unknown coefficients. Using the initial value conditions, it follows that a=1,b=0.  (uψ+uψ+2uψ)+2x(uψ+uψ)+(x2+2)uψ=ex22(ψ+ψ) It follows that ex22(ψ+ψ)=0 which gives ψ+ψ=0  ψ(x)=acosx+bsinx, where a and b are unknown coefficients. Using the initial value conditions, it follows that a=1,b=0.

Cason Harmon

Cason Harmon

Beginner2022-03-23Added 8 answers

Step 1
Consider ϕ=uψ. Then since u=xu we also get
u =uxu=u+x2u. Thus ϕ=(xu)ψ+uψ=u(ψxψ) and ϕ =u(x21)ψ+(xu)ψ+(xu)ψ+uψ =u((x21)ψ2xψ+ψ ).
Therefore, 0=u((x2-1)ψ-2xψ'+ψ'')+2xu(ψ'-xψ)+(x2+2)uψ=u(ψ(x2-1-2x2+x2+2)+

+ψ'(-2x+2x)+ψ'')=u(ψ+ψ'')
Step 2
Now if u is not the trivial solution u0 we can thus divide be u to get
ψ +ψ=0
the solution of which is of the form asin(x)+bcos(x). Now it suffices to see that a non trivial solution to y=xy cannot be periodic (if x<0 the sign of y' is equal to the sign of y).
Thus the only options are B,D. This means ψ(x)=cos(x), and thus either u=ex22. By differentiating this we can see that only B satisfies the condition u=xu.

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