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 toA) (cos2(x))e−x22B) (cos⁡(x))e−x22C) (1+x2)e−x22D) (cos⁡(x))e−x2

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Chad Robinson · Accepted Answer

Step 1It&#039;s almost tempting to use reduction of order here but unfortunately that doesn&#039;t work here as u is not a solution of&amp;nbsp;y&amp;nbsp;″+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 &#039;hope&#039; of cancellation of some terms. You are almost there.Step 2Consider&amp;nbsp;u=e−x22&amp;nbsp;(just taking&amp;nbsp;c=1&amp;nbsp;in u that you obtained and this can be done due to homogeneity of the blue colored ODE). Then note that&amp;nbsp;u′=−xe−x22&amp;nbsp;and&amp;nbsp;u&amp;nbsp;″=(x2−1)e−x22. Substitute&amp;nbsp;ϕ=uψ&amp;nbsp;in&amp;nbsp;y&amp;nbsp;″+2xy′+(x2+2)y=0&amp;nbsp;to get&amp;nbsp;&amp;nbsp;(u″ψ+uψ″+2u′ψ′)+2x(uψ′+u′ψ′)+(x2+2)uψ=e−x22(ψ″+ψ)&amp;nbsp;It follows that&amp;nbsp;e−x22(ψ″+ψ)=0&amp;nbsp;which gives&amp;nbsp;ψ″+ψ=0&amp;nbsp;⟹&amp;nbsp;ψ(x)=acos⁡x+bsin⁡x,&amp;nbsp;where&amp;nbsp;a&amp;nbsp;and&amp;nbsp;b&amp;nbsp;are unknown coefficients.&amp;nbsp;Using the initial value conditions, it follows that&amp;nbsp;a=1,b=0.&amp;nbsp;&amp;nbsp;(u″ψ+uψ″+2u′ψ′)+2x(uψ′+u′ψ′)+(x2+2)uψ=e−x22(ψ″+ψ)&amp;nbsp;It follows that&amp;nbsp;e−x22(ψ″+ψ)=0&amp;nbsp;which gives&amp;nbsp;ψ″+ψ=0&amp;nbsp;⟹&amp;nbsp;ψ(x)=acos⁡x+bsin⁡x,&amp;nbsp;where&amp;nbsp;a&amp;nbsp;and&amp;nbsp;b&amp;nbsp;are unknown coefficients.&amp;nbsp;Using the initial value conditions, it follows that&amp;nbsp;a=1,b=0.

Cason Harmon · Accepted Answer

Step 1Consider&amp;nbsp;ϕ=uψ. Then since&amp;nbsp;u′=−xu&amp;nbsp;we also getu&amp;nbsp;″=−u−xu′=−u+x2u. Thus&amp;nbsp;ϕ′=(−xu)ψ+uψ′=u(ψ′−xψ)&amp;nbsp;and&amp;nbsp;ϕ&amp;nbsp;″=u(x2−1)ψ+(−xu)ψ′+(−xu)ψ′+uψ&amp;nbsp;″=u((x2−1)ψ−2xψ′+ψ&amp;nbsp;″).Therefore,&amp;nbsp;0=u((x2-1)ψ-2xψ&#039;+ψ&#039;&#039;)+2xu(ψ&#039;-xψ)+(x2+2)uψ=u(ψ(x2-1-2x2+x2+2)++ψ&#039;(-2x+2x)+ψ&#039;&#039;)=u(ψ+ψ&#039;&#039;)Step 2Now if u is not the trivial solution&amp;nbsp;u≡0&amp;nbsp;we can thus divide be u to getψ&amp;nbsp;″+ψ=0the solution of which is of the form&amp;nbsp;asin⁡(x)+bcos⁡(x). Now it suffices to see that a non trivial solution to&amp;nbsp;y′=−xy&amp;nbsp;cannot be periodic (if&amp;nbsp;x&amp;lt;0&amp;nbsp;the sign of y&#039; is equal to the sign of y).Thus the only options are B,D. This means&amp;nbsp;ψ(x)=cos⁡(x), and thus either&amp;nbsp;u=e−x22. By differentiating this we can see that only B satisfies the condition&amp;nbsp;u′=−xu.

Let u be a solution of the differential

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