Find the general solution of y'' - 6y' + 9y = 36√(x)…

Find the general solution of $y'' - 6y' + 9y = 36\sqrt{x} - \frac{12x+1}{x\sqrt{x}}$.

Step 1: Rewriting the forcing term as powers of $x$

Split the fraction over $x\sqrt{x}=x^{3/2}$: \[\frac{12x+1}{x\sqrt{x}} = \frac{12x}{x^{3/2}} + \frac{1}{x^{3/2}} = 12x^{-1/2} + x^{-3/2}.\]

So the equation reads

\[ y'' - 6y' + 9y = 36x^{1/2} - 12x^{-1/2} - x^{-3/2}. \]

Step 2: Homogeneous solution

The characteristic equation is

\[ r^2-6r+9 = (r-3)^2 = 0 \;\Longrightarrow\; r=3\ (\text{double}), \]

so $y_h = (C_1+C_2x)e^{3x}$.

Step 3: Particular solution — a power-function ansatz

The right side is built from $x^{1/2}$ and the powers one and two derivatives below it, $x^{-1/2}$ and $x^{-3/2}$ — exactly the powers generated by differentiating $x^{1/2}$ twice. This suggests the ansatz

\[ y_p = c\,x^{1/2} + d\,x^{-1/2}. \]

Its derivatives are

\[ y_p' = \frac{c}{2}x^{-1/2} - \frac{d}{2}x^{-3/2}, \qquad y_p'' = -\frac{c}{4}x^{-3/2} + \frac{3d}{4}x^{-5/2}. \]

Substituting into $y_p''-6y_p'+9y_p$ and collecting by power of $x$: \[y_p''-6y_p'+9y_p = \frac{3d}{4}x^{-5/2} + \left(-\frac{c}{4}+3d\right)x^{-3/2} + (-3c+9d)x^{-1/2} + 9c\,x^{1/2}.\]

The right-hand side $36x^{1/2}-12x^{-1/2}-x^{-3/2}$ has no $x^{-5/2}$ term, forcing $d=0$. With $d=0$ the remaining terms are $-\tfrac{c}{4}x^{-3/2}-3c\,x^{-1/2}+9c\,x^{1/2}$; matching the $x^{1/2}$ coefficient gives $9c=36$, i.e. $c=4$. This single value also matches the other two powers automatically: $-\tfrac{c}{4}=-1$ and $-3c=-12$, exactly as required. Hence

\[ y_p = 4\sqrt{x}. \]

Step 4: General solution

The solution is recorded in the result below.

Step 5: Verification

With $y_p=4x^{1/2}$, $y_p'=2x^{-1/2}=\dfrac{2}{\sqrt{x}}$ and $y_p''=-x^{-3/2}=-\dfrac{1}{x\sqrt{x}}$. Then

\[ y_p''-6y_p'+9y_p = -\frac{1}{x\sqrt{x}} - \frac{12}{\sqrt{x}} + 36\sqrt{x}. \]

Writing $\dfrac{12}{\sqrt{x}}=\dfrac{12x}{x\sqrt{x}}$ so both fractions share the denominator $x\sqrt{x}$, \[-\frac{1}{x\sqrt{x}} - \frac{12x}{x\sqrt{x}} = -\frac{12x+1}{x\sqrt{x}},\]

so $y_p''-6y_p'+9y_p = 36\sqrt{x} - \dfrac{12x+1}{x\sqrt{x}}$, matching the right side exactly.\ $\checkmark$ The homogeneous part $(C_1+C_2x)e^{3x}$ satisfies the associated homogeneous equation by construction (fundamental system at the double root $r=3$), so by linearity the boxed expression solves the full equation for all $C_1,C_2$.