Question

Suppose $$ y_{p}=\bar{y}+a_{1} y_{1}+a_{2} y_{2} $$ is a particular solution of $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F(x), $$ where \(y_{1}\) and \(y_{2}\) are solutions of the complementary equation $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=0. $$ Show that \(\bar{y}\) is also a solution of \((\mathrm{A})\)

Short answer

Question: Show that if $y_p = \bar{y} + a_1y_1 + a_2y_2$, where $\bar{y}$ is a particular solution, and $y_1$ and $y_2$ are solutions to the complementary equation with the same coefficients as the original equation, then $\bar{y}$ is also a solution of the original equation. Answer: We can show that $\bar{y}$ is a solution to the original equation by following these steps: 1. Substitute the given \(y_p\) into the original equation. 2. Find the first and second derivatives of \(y_p\) with respect to \(x\). 3. Substitute the expressions for the first and second derivatives of \(y_p\) into the original equation. 4. Use the fact that \(y_1\) and \(y_2\) are solutions to the complementary equation. 5. Combine the resulting equations. Following these steps, we can see that the equation involving $\bar{y}$ has the form $P_0(x)\bar{y}'' + P_1(x)\bar{y}' + P_2(x)\bar{y} = F(x)$. This confirms that $\bar{y}$ is a solution to the original equation.

Step-by-step solution

Substitute the particular solution into the original equation

Substitute \(y_p = \bar{y} + a_1y_1 + a_2y_2\) into the original equation: $$ P_0(x)(\bar{y} + a_1y_1 + a_2y_2)^{\prime\prime} + P_1(x)(\bar{y} + a_1y_1 + a_2y_2)^{\prime} + P_2(x)(\bar{y} + a_1y_1 + a_2y_2) = F(x). $$

Find the derivatives of the particular solution

Find the first derivative of \(y_p\) with respect to \(x\): $$ y_p' = \bar{y}' + a_1y_1' + a_2y_2', $$ and the second derivative of \(y_p\) with respect to \(x\): $$ y_p'' = \bar{y}'' + a_1y_1'' + a_2y_2''. $$

Substitute the derivatives into the equation

Substitute the expressions for the first and second derivatives of \(y_p\) into the original equation: $$ P_0(x)(\bar{y}'' + a_1y_1'' + a_2y_2'') + P_1(x)(\bar{y}' + a_1y_1' + a_2y_2') + P_2(x)(\bar{y} + a_1y_1 + a_2y_2) = F(x). $$

Use the fact that \(y_1\) and \(y_2\) are solutions to the complementary equation

Since \(y_1\) and \(y_2\) are solutions to the complementary equation, we know that: $$ P_0(x)y_1'' + P_1(x)y_1' + P_2(x)y_1 = 0 $$ and $$ P_0(x)y_2'' + P_1(x)y_2' + P_2(x)y_2 = 0. $$ Multiply the first equation by \(a_1\) and the second equation by \(a_2\): $$ a_1[P_0(x)y_1'' + P_1(x)y_1' + P_2(x)y_1] = 0 $$ and $$ a_2[P_0(x)y_2'' + P_1(x)y_2' + P_2(x)y_2] = 0. $$

Combine equations

Add the two equations from Step 4 together, along with the equation from Step 3: $$ P_0(x)\bar{y}'' + P_1(x)\bar{y}' + P_2(x)\bar{y} = F(x). $$ The above equation shows that \(\bar{y}\) is a solution to the original equation, just as we wanted to prove.

Key concepts

Particular Solution

Many physical phenomena are modeled using differential equations with conditions specified at the boundaries of the domain, known as boundary value problems (BVPs). These problems differ from initial value problems, which provide a condition at a single point, usually the start of the interval. Boundary value problems require solutions to satisfy conditions at two or more points.

For example, to model the temperature distribution along a rod, conditions such as the temperature at both ends of the rod would serve as the boundary values. In this scenario, the solution must comply with both of these conditions, creating a problem that is often more complex and may have more than one solution or sometimes no solution at all.

The essence of solving a BVP lies in finding a function that not only satisfies the given differential equation but also meets the prescribed boundary conditions. These conditions act as crucial constraints that narrow down the infinitely many solutions of a differential equation to those of interest for the particular physical situation at hand.

When solving for a particular solution in the context of BVPs, it's instrumental to first solve the complementary equation to understand the general solution's structure. Then the additional information from the boundary conditions is used to find the specific values for the solution. Consequently, these problems are central in fields such as engineering, physics, and applied mathematics as they accurately describe the real-world constraints of a system or phenomenon.