Solve the differential equation using variation of parameters: $y^{″}+3y^{\prime }+2y=15+e^x$. Which of the following is the general solution?

• A. $y=C_1{e}^{-x}+C_2{e}^{-2x}+\frac{15}{2}+\frac{e^x}{6}$
• B. $y=C_1{e}^x+C_2{e}^{2x}+\frac{15}{2}+\frac{e^x}{2}$
• C. $y=C_1{e}^{2x}+C_2{e}^{-3x}+15x+e^x$
• D. $y=C_1{e}^{-x}+C_2{e}^{-2x}+15x+e^x$

Answer: (A)

Verified step by step guidance

1. Step 1: Identify the type of differential equation. The given equation is a second-order linear non-homogeneous differential equation: $y\text{'}\text{'}\; +\; 3y\text{'}\; +\; 2y\; =\; 15\; +\; e^x$. The solution will consist of the complementary solution (solution to the homogeneous equation) and a particular solution (solution to the non-homogeneous part).
2. Step 2: Solve the homogeneous equation. The homogeneous equation is $y\text{'}\text{'}\; +\; 3y\text{'}\; +\; 2y\; =\; 0$. Solve this by finding the characteristic equation: $r^2\; +\; 3r\; +\; 2\; =\; 0$. Factorize the characteristic equation to find the roots, which will determine the complementary solution.
3. Step 3: Write the complementary solution. Based on the roots of the characteristic equation, the complementary solution will be of the form $y\_c\; =\; C\_1\; e^\{-x\}\; +\; C\_2\; e^\{-2x\}$, where $C\_1$ and $C\_2$ are constants.
4. Step 4: Find the particular solution using variation of parameters. For the non-homogeneous term $15\; +\; e^x$, assume a particular solution of the form $y\_p\; =\; A\; +\; B\; e^x$. Substitute $y\_p$ into the original equation and solve for $A$ and $B$ by equating coefficients.
5. Step 5: Combine the complementary and particular solutions. The general solution to the differential equation is $y\; =\; y\_c\; +\; y\_p$, which combines the complementary solution $y\_c\; =\; C\_1\; e^\{-x\}\; +\; C\_2\; e^\{-2x\}$ and the particular solution $y\_p$ obtained in Step 4.