The Particular solution to a Euler/Cauchy Equation

Question

This is the same problem as in my most recent question. I'll be picking up where that left off.

The given is: $$x^2y^{"}+xy^{'}+y=\ln x$$ The Aux. equation is: $$y=C_1\cos(\ln x)+C_2\sin(\ln x)$$ Now we can see that $y_1=\cos(\ln x)$ and $y_2=\sin(\ln x)$ and taking the wronskian of these two values and the given formula in standard form gives us: $$W=\begin{bmatrix}\cos(\ln x)\ \sin(\ln x)\\ -\frac{\sin(\ln x)}{x}\ \frac{\cos(\ln x)}{x} \end{bmatrix}=\frac{\cos^2(\ln x)}{x}+\frac{\sin^2(\ln x)}{x}=\frac{1}{x}$$ $$W_1=\frac{\ln(x)\cos(\ln x)}{x^2}$$ $$W_2=\frac{\ln(x)\sin(\ln x)}{x^2}$$ Now finding $U_{1,2}^{'}$ is just $\frac{W_{n}}{W}$: $$U_1^{'}=\frac{\ln(x)\cos(\ln x)}{x}$$ $$U_2^{'}=\frac{\ln(x)\sin(\ln x)}{x}$$ Then after integrating I have: $$U_1=\ln(x)\sin(\ln x)+\cos(\ln x)$$ $$U_2=-\ln(x)\cos(\ln x)+\sin(\ln x)$$ The rest of the problem is easy enough, I just wanna make sure my work is correct before solving.

Answer 1

In the original question, there was an answer with the hint to substitute $x=e^t$. The equation is then written as $$y''(t)+y(t)=t$$ Hopefully you found $y_h(t)=a\cos(t)+b\sin(t)$ for the homogeneous equation. Here you find a particular solution $y_p(t)=At+B$ with $A=1$ and $B=0$.

Substitute back, then $y_h(x)=a\cos(\ln(x))+b\sin(\ln(x))$ and a particular solution is then given by $y_p(x)=\ln(x)$.

Finally $y(x)=y_h(x)+y_p(x)$.

Answer 2

By the method of undetermined coefficients, the trial solution for the right side $x^0\ln x$ is $$ y_p(x)=A+B\ln x. $$ Inserting gives $y_p'(x)=Bx^{-1}$ and $y_p''(x)=-x^{-2}$ so that $$ -Bx^{2-2}+Bx^{1-1}+[A+B\ln x]=\ln x\\ \implies A=0,~~B=1, \\ y_p(x)=\ln x $$ This is, as far as I can see, not conform with your calculation.

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Let's check. The system to solve for the variation of constants is $$ \begin{bmatrix}y_1&y_2\\y_1'&y_2'\end{bmatrix} \begin{bmatrix}U_1'\\U_2'\end{bmatrix} = \begin{bmatrix}0\\f/x^2\end{bmatrix} \implies \begin{bmatrix}U_1'\\U_2'\end{bmatrix} = \frac1{x^2W} \begin{bmatrix}y_2'&-y_2\\-y_1'&y_1\end{bmatrix} \begin{bmatrix}0\\f\end{bmatrix} = \frac1{x^2W}\begin{bmatrix}-y_2f\\y_1f\end{bmatrix} $$ And thus with $\frac1{x^2W}=\frac1x$ one should get $U_1'=-\frac{\ln x}{x}\sin(\ln x)$, $U_2'=\frac{\ln x}{x}\cos(\ln x)$. This is rotated by $90°$ from what you calculated.

Answer 3

Note that $$U_1(x)=-\int \frac{\ln x \sin(\ln x)}{w} dx+D_1=-\int x \ln x \sin(\ln x)dx+D_1$$ $$U_2(x)=\int \frac{\ln x \cos(\ln x)}{w} dx+D_2=\int x \ln x \cos(\ln x)dx+D_2$$ then the total solution would be $$y(x)=U_1(x) \cos(\ln x)+U_2(x) \sin(\ln x)$$

Answer 4

The wronskian is correct: $$W=\begin{bmatrix}\cos(\ln x)\ \sin(\ln x)\\ -\frac{\sin(\ln x)}{x}\ \frac{\cos(\ln x)}{x} \end{bmatrix}=\frac{1}{x}$$

But you should have: $$W_1={\ln(x)\cos(\ln x)}$$ $$W_2=-{\ln(x)\sin(\ln x)}$$ And also : $$U_1'=\dfrac {W_1}{W}=xW_1=x{\ln(x)\cos(\ln x)}$$ $$U_2'=\dfrac {W_2}{W}=xW_2=-x{\ln(x)\sin(\ln x)}$$

The integrals are complicated to evaluate but in the end the result is simple. The Method of Undetermined Parameters is far better for this differential equation.