Question

Consider the differential equation \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\). In each exercise, the complementary solution, \(y_{C}(t)\), and nonhomogeneous term, \(g(t)\), are given. Determine \(\alpha\) and \(\beta\) and then find the general solution of the differential equation. $$ y_{C}(t)=c_{1}+c_{2} e^{-t}, \quad g(t)=t $$

Short answer

Based on the given information, we were able to find the general solution of the given differential equation as \(y(t) = c_1 + c_2e^{-t} + t\). The constants \(\alpha\) and \(\beta\) were determined to be 1 and 0, respectively. To find the general solution, we identified the linearly independent solutions, differentiated the complementary solution, plugged the results into the given differential equation, found the particular solution, and added the complementary and particular solutions together.

Step-by-step solution

Identify the linearly independent solutions from the complementary solution given

Two functions are linearly independent if each function is not a product of the other with a constant. Since the complementary function \(y_C(t) = c_1 + c_2e^{-t}\) is given, we can conclude that the linearly independent solutions are $$ y_1(t) = 1 \\ y_2(t) = e^{-t} $$

Differentiate the given complementary solution twice

Differentiate \(y_C(t)=c_1+c_2e^{-t}\) with respect to t twice to get \(y_C'(t)\) and \(y_C''(t)\). $$ y_C'(t) = -c_2 e^{-t}\\ y_C''(t) = c_2 e^{-t} $$

Plug the derived results into the given differential equation and solve for \(\alpha\) and \(\beta\)

Plug \(y_C''(t)\), \(y_C'(t)\), and \(y_C(t)\) into the given differential equation \(y''+\alpha y' + \beta y = g(t)\) to determine \(\alpha\) and \(\beta\). $$ y_C''(t) + \alpha y_C'(t) + \beta y_C(t) = c_2e^{-t} - \alpha c_2e^{-t} + \beta(c_1 + c_2e^{-t}) = t $$ Observe that \(g(t) = t \neq ce^{-t}\) for any constants \(c\), therefore the left-hand side should contain a term that contains just t. In order to make this happen, we can set \(\beta\) to zero. $$ \beta = 0 $$ Now let's find \(\alpha\) by considering the term containing \(e^{-t}\): $$ (c_2 - \alpha c_2) e^{-t} = 0 \implies \alpha = 1 $$ Hence, we have \(\alpha = 1\) and \(\beta = 0\). Now, the given differential equation becomes: $$ y''+ y' = t $$

Find the particular solution \(y_p(t)\)

Since \(y_C(t)\) contains only constants and exponential functions, we can try the polynomial function of degree 1 for \(y_p(t)\): $$ y_p(t) = At + B $$ Differentiate \(y_p(t)\) with respect to t twice to obtain: $$ y_p'(t) = A \\ y_p''(t) = 0 $$ Now, plug these results into the modified differential equation: $$ y_p''(t) + y_p'(t) = 0 + A = t $$ Comparing the coefficients, we get: $$ A = 1 \\ B = 0 $$ Therefore, the particular solution is given by: $$ y_p(t) = t $$

Find the general solution \(y(t)\)

Now that we have both the complementary solution \(y_C(t)\) and the particular solution \(y_p(t)\), we can find the general solution by adding them: $$ y(t) = y_C(t) + y_p(t) = (c_1 + c_2e^{-t}) + t $$ So the general solution of the given differential equation is: $$ y(t) = c_1 + c_2e^{-t} + t $$

Key concepts

Complementary Solution

The complementary solution, often denoted as y_C(t), plays a crucial role in solving nonhomogeneous differential equations. It represents the general solution to the associated homogeneous equation, which is obtained when the nonhomogeneous term g(t) is set to zero. In simple terms, it encompasses all possible solutions that arise from the equation without considering any external forces or inputs.

In the provided exercise, the complementary solution is y_C(t) = c₁ + c₂e^-t. This tells us that any linear combination of c₁ and c₂e^-t will satisfy the homogeneous differential equation. Importantly, we identify that the functions 1 and e^-t are linearly independent, meaning they form a basis for the solution space of the corresponding homogeneous equation.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation includes a non-zero term g(t) which makes the equation nonhomogeneous or inharmonious. This is because the presence of g(t), often referred to as the forcing function, introduces an additional component to the overall solution that must specifically address this nonhomogeneous part.

In the exercise, the nonhomogeneous term is simply t, making the differential equation take the form y'' + \(alpha\)y' + \(beta\)y = t. To solve such an equation, we need to find not only the complementary solution, which we have, but also a particular solution that will work for the nonhomogeneous part. By combining these two pieces, we can construct the general solution to the full nonhomogeneous differential equation.

Particular Solution

The particular solution of a nonhomogeneous differential equation is a specific solution that accounts for the presence of the nonhomogeneous term g(t). Unlike the complementary solution, which addresses the associated homogeneous equation, the particular solution is crafted to only fit the part that is affected by g(t).

In this exercise, we looked for a particular solution in the form of y_p(t) = At + B, a polynomial of the same degree as g(t). The choice of this form is motivated by the method of undetermined coefficients, which allows us to find a particular solution by differentiating this assumed y_p(t) and equating the derivatives to match the nonhomogeneous part of the differential equation. After this process, coefficients A and B are determined, yielding the particular solution.

General Solution

The general solution of a differential equation encapsulates all possible solutions within it. It is the sum of the complementary solution, which solves the homogeneous part, and the particular solution, which directly addresses the nonhomogeneous part. Therefore, the general solution is not just a single function but rather a family of functions that adheres to all the conditions set by the original differential equation.

Following the logic of the step-by-step solution, once we have both the complementary solution, y_C(t), and the particular solution, y_p(t), we combine them to find the general solution: y(t) = y_C(t) + y_p(t). In this case, the general solution to the nonhomogeneous differential equation is y(t) = c₁ + c₂e^-t + t, which encompasses the effects of both the homogeneous properties of the equation and the specific influence of the nonhomogeneous term.