Find the general solution to x^2y" - 3xy' + 4y = 0.

Question

Find the general solution to

[latex]x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=0 \quad\quad\quad\quad\quad (6)[/latex].

Accepted Answer

The characteristic equation is

[latex]\lambda^{2}-4 \lambda+4=(\lambda-2)^{2}=0[/latex]

with the single root [latex]\lambda=2[/latex]. Thus one solution is [latex]y_{1}(x)=x^{2}[/latex]. To find a second solution, we use formula (8) on page 667. First, we write (6) in the form

[latex]y_{2}=v y_{1}=y_{1}(x) \int \frac{e^{-\int a(x) d x}}{y_{1}^{2}(x)} d x[/latex]

[latex]y^{\prime \prime}-\frac{3}{x} y^{\prime}+\frac{4}{x^{2}} y=0 .[/latex]

As in Section 11.7, [latex]a(x)=-\frac{3}{x}[/latex] so that

[latex]e^{-\int a(x) d x}=e^{\int 3 / x d x}=e^{3 \ln x}=x^{3}[/latex]

and

[latex]y_{2}(x)=y_{1}(x) \int \frac{e^{-\int a(x) d x}}{y_{1}^{2}(x)} d x=x^{2} \int \frac{x^{3}}{x^{4}} d x=x^{2} \ln x[/latex].

Thus the general solution to equation (6) is

[latex]y(x)=c_{1} x^{2}+c_{2} x^{2} \ln x=x^{2}\left(c_{1}+c_{2} \ln x\right)[/latex].

Question 11.12.2: Find the general solution to x^2y" - 3xy' + 4y = 0.

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