Find the general solution to x^2y" + 2xy' - 12y = √x.

Question

Find the general solution to

[latex]x^{2} y^{\prime \prime}+2 x y^{\prime}-12 y=\sqrt{x} \quad\quad\quad\quad\quad (9)[/latex].

Accepted Answer

In Example 1 we found the homogeneous solutions

[latex]y_{1}=x^{-4} \quad ext { and } \quad y_{2}=x^{3}[/latex].

Then

[latex]W\left(y_{1}, y_{2} ight)(x)=\left|\begin{array}{cc}x^{-4} & x^{3} \-4 x^{-5} & 3 x^{2}\end{array} ight|=3 x^{-2}+4 x^{-2}=\frac{7}{x^{2}} ext { and } \frac{1}{W}=\frac{x^{2}}{7}[/latex]

We rewrite (9) in the standard form:

[latex]y^{\prime \prime}+\frac{1}{x} y^{\prime}-\frac{12}{x^{2}} y=x^{-3 / 2}[/latex]

This is the form for which the variation of constants formulas, formulas (7) and (8), on page 684 apply. Then [latex]f(x)=x^{-3 / 2}[/latex], and we obtain

[latex]\begin{aligned}&c_{1}^{\prime}(x)=\frac{-f(x) y_{2}(x)}{y_{1}(x) y_{2}^{\prime}(x)-y_{1}^{\prime}(x) y_{2}(x)}=\frac{-f(x) y_{2}(x)}{W\left(y_{1}, y_{2} ight)(x)} \&c_{2}^{\prime}(x)=\frac{f(x) y_{1}(x)}{y_{1}(x) y_{2}^{\prime}(x)-y_{1}^{\prime}(x) y_{2}(x)}=\frac{f(x) y_{1}(x)}{W\left(y_{1}, y_{2} ight)(x)}\end{aligned}[/latex]

[latex]c_{1}^{\prime}(x)=\frac{-f(x) y_{2}(x)}{W\left(y_{1}, y_{2} ight)(x)}=\frac{x^{2}}{7}\left(-x^{-3 / 2} ight)\left(x^{3} ight)=\frac{-x^{7 / 2}}{7}[/latex]

and

[latex]c_{2}^{\prime}(x)=\frac{f(x) y_{1}(x)}{W\left(y_{1}, y_{2} ight)(x)}=\frac{x^{2}}{7}\left(x^{-3 / 2} ight)\left(x^{-4} ight)=\frac{1}{7} x^{-7 / 2}[/latex]

Hence

[latex]c_{1}(x)=-\frac{1}{7} \cdot \frac{2}{9} x^{9 / 2}, \quad c_{2}(x)=-\frac{1}{7} \cdot \frac{2}{5} x^{-5 / 2}[/latex]

so that

[latex]\begin{aligned}y_{p}(x) &=c_{1}(x) y_{1}(x)+c_{2}(x) y_{2}(x)=-\frac{1}{7}\left[\frac{2}{9} x^{9 / 2} \cdot x^{-4}+\frac{2}{5} x^{-5 / 2} \cdot x^{3} ight] \&=\frac{-x^{1 / 2}}{7}\left(\frac{2}{9}+\frac{2}{5} ight)=-\frac{4}{45} x^{1 / 2}\end{aligned}[/latex]

Thus the general solution is given by

[latex]y(x)=c_{1} x^{-4}+c_{2} x^{3}-\frac{4}{45} x^{1 / 2}[/latex]

Question 11.12.4: Find the general solution to x^2y" + 2xy' - 12y = √x.

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