Question

Let \(|G(t)|T>0\) and for some constants \(M\) and \(a\) in $$ Y^{\prime}=A Y+G $$ Prove that for some constants \(K\) and \(b\), $$ |\Phi(t)|

Short answer

Question: Prove that there exists constants \(K\) and \(b\) such that \(|\Phi(t)| < K e^{b t}\), given the inequality \(|G(t)| T > 0\) with some constants \(M\) and \(a\) and the differential equation \(Y^{\prime} = A Y + G\) where \(A\) is a constant. Solution: We have shown that by applying the variation of parameters method on the given differential equation and using the integral bound, we can find two constants \(K = \frac{M}{|C|(a-2A)}\) and \(b = a-A\) such that \(|\Phi(t)|<K e^{b t}\).

Step-by-step solution

Solve the homogeneous part

For the homogeneous part of the given differential equation \(Y^{\prime} = A Y + G\), we will solve the equation \(Y^{\prime} = A Y\). This is a first-order linear homogeneous differential equation and its general solution will be in the form of \(Y_H = C e^{A t}\), where \(C\) is the constant of integration.

Apply Variation of Parameters

To find the particular solution \(Y_P\), apply the variation of parameters method which is given by: $$Y_P(t) = -Y_H(t)\int \frac{G(t)}{Y_H^2(t)} dt$$ Here \(Y_H(t) = C e^{A t}\). Now, plug this into the expression for \(Y_P\): $$Y_P(t) = -C e^{A t}\int \frac{G(t)}{(C e^{A t})^2} dt = -\frac{1}{C e^{A t}}\int G(t) e^{-2 A t} dt$$

Combine homogeneous and particular solutions

The general solution of the given differential equation is the sum of homogenous \(Y_H(t)\) and particular \(Y_P(t)\) solutions: $$Y(t) = Y_H(t) + Y_P(t) = C e^{A t} -\frac{1}{C e^{A t}}\int G(t) e^{-2 A t} dt$$ We need to find the expression for \(\Phi(t)\) which is defined as: \(\Phi(t) = Y(t) - C e^{A t}\). So, subtracting \(C e^{A t}\) from \(Y(t)\), we get: $$\Phi(t)=-\frac{1}{C e^{A t}}\int G(t) e^{-2 A t} dt$$

Find appropriate constants \(K\) and \(b\)

Now, we need to determine the appropriate values for the constants \(K\) and \(b\) such that \(|\Phi(t)| < K e^{b t}\) is satisfied. Notice the given condition for \(|G(t)|\). Utilizing the integral bound, we have: $$\int |G(t)| e^{-2At} dt \leq \int M e^{(a - 2A)t} dt$$ Integrating with respect to \(t\), we obtain: $$\int G(t) e^{-2At} dt \leq \frac{Me^{(a-2A)t}}{a-2A}$$ Recall the expression for \(\Phi(t)\). We can make the following substitutions: - \(K = \frac{M}{|C|(a-2A)}\) - \(b = a-A\) With these constants, we have $$\begin{align*} |\Phi(t)| &= \frac{1}{|C| e^{A t}}\int |G(t)| e^{-2 A t} dt \\ & \leq \frac{1}{|C| e^{A t}}\int M e^{(a - 2A)t} dt \\ & = \frac{M}{|C|(a-2A)} e^{(a - A)t} \\ &= K e^{bt} \end{align*}$$ Thus, we have proven that \(|\Phi(t)|<K e^{b t}\) for some constants \(K\) and \(b\).