Find the general solution of y'' - 5y' + 6y = x

Find the general solution of $y'' - 5y' + 6y = x$.

Step 1: Homogeneous solution

\[ r^2-5r+6=0 \implies (r-2)(r-3)=0 \implies r_1=2,\ r_2=3. \]

\[ Y_h = C_1e^{2x} + C_2e^{3x}. \]

Step 2: Particular solution

The right-hand side $f(x)=x$ is a polynomial of degree 1, and $r=0$ is not a root of the characteristic equation, so try $y^*=Ax+B$: $y^{*\prime}=A, \qquad y^{*\prime\prime}=0.$

Substitute into the equation: $0 - 5A + 6(Ax+B) = x \implies 6Ax + (6B-5A) = x.$

Step 3: Match coefficients

\[ x^1:\ 6A=1 \implies A=\frac{1}{6}, \qquad x^0:\ 6B-5A=0 \implies B=\frac{5A}{6}=\frac{5}{36}. \]

\[ y^* = \frac{x}{6} + \frac{5}{36}. \]

Step 4: Verify

\[ y^{*\prime\prime} - 5y^{*\prime} + 6y^* = 0 - 5\cdot\frac{1}{6} + 6\!\left(\frac{x}{6}+\frac{5}{36}\right) = -\frac{5}{6}+x+\frac{5}{6} = x.\ \checkmark \]

Step 5: General solution

The solution is recorded in the result below.