Given that y_1=2x and y_2=(x+1)^2 are two particular…

Given that $y_1=2x$ and $y_2=(x+1)^2$ are two particular solutions, find the general solution of $(3x^2+x)y'' - (3x+1)y' = -(6x+2)$.

Step 1: Factoring to a canonical form

Since $3x^2+x=x(3x+1)$ and $-(6x+2)=-2(3x+1)$, the equation reads

\[ x(3x+1)y'' - (3x+1)y' = -2(3x+1). \]

Every term carries the common factor $(3x+1)$. Dividing it out (for $x\neq-1/3$) gives the much simpler canonical form

\[ xy'' - y' = -2. \]

Step 2: Verifying $y_1$ and $y_2$ in the reduced equation

For $y_1=2x$: $y_1'=2$, $y_1''=0$, so $xy_1''-y_1' = x\cdot0-2=-2.\ \checkmark$

For $y_2=(x+1)^2$: $y_2'=2(x+1)$, $y_2''=2$, so $xy_2''-y_2' = 2x-2(x+1) = -2.\ \checkmark$

Both are indeed particular solutions of $xy''-y'=-2$ (and hence, after multiplying back by $(3x+1)$, of the original equation).

Step 3: Solving the associated homogeneous equation

Set $p=y'$ in $xy''-y'=0$, so $xp'-p=0$, a separable first-order equation: \[\frac{dp}{p} = \frac{dx}{x} \;\Longrightarrow\; p = C_1x.\]

Integrating $y'=C_1x$ once more and renaming $C_1/2\to C_1$: $y_h = C_1x^2 + C_2$.

so $\{x^2,\,1\}$ is a fundamental system for the homogeneous equation $xy''-y'=0$.

Step 4: Confirming the fundamental system

Directly: $y=x^2$ gives $x(2)-2x=0\ \checkmark$; $y=1$ gives $x(0)-0=0\ \checkmark$. Their Wronskian is

\[ W(x^2,1) = x^2\cdot0 - 2x\cdot1 = -2x \neq 0 \quad (x\neq0), \]

confirming linear independence away from $x=0$.

Note also that $y_2-y_1 = (x+1)^2-2x = x^2+1$ is itself a homogeneous solution (check: $x(2)-2x=0\ \checkmark$), consistent with $\text{span}\{x^2,1\}=\text{span}\{x^2+1,1\}$ — the two given particular solutions differ by exactly a homogeneous combination, as they must.

Step 5: General solution of the original equation

Using $y_1=2x$ as the particular solution and $y_h=C_1x^2+C_2$ as the homogeneous general solution:

Step 6: Verification in the original (unreduced) equation

With $y=C_1x^2+C_2+2x$: $y'=2C_1x+2$, $y''=2C_1$. Substitute into the original left side: $(3x^2+x)y''-(3x+1)y' = x(3x+1)(2C_1) - (3x+1)(2C_1x+2)$.

Factor out $(3x+1)$: \[= (3x+1)\bigl[2C_1x - 2C_1x - 2\bigr] = (3x+1)(-2) = -(6x+2).\ \checkmark\]

The $C_1x$ terms cancel exactly — a direct consequence of $y=C_1x^2+\cdots$ solving the homogeneous equation — and the result matches the right side identically for all $C_1,C_2$.