Question

(a) Show that $$\begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned}$$ and so on; that is, for any positive integral \(n\) $$D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y$$ Thus show that if \(L(D)\) is any polynomial in the operator \(D,\) then $$L(D)\left(e^{a x} y\right)=e^{a x} L(D+a) y$$ This is called the exponential shift. (b) Use (a) to show that $$\begin{aligned} (D-1)^{3}\left(e^{x} y\right) &=e^{x} D^{3} y \\ \left(D^{2}+D-6\right)\left(e^{-3 x} y\right) &=e^{-3 x}\left(D^{2}-5 D\right) y \end{aligned}$$ (c) Replace \(D\) by \(D-a,\) to obtain $$e^{a x} P(D) y=P(D-a) e^{a x} y$$ This is called the inverse exponential shift. (d) Using (c), we can change a differential equation whose right-hand side is an exponential times a polynomial, to one whose right-hand side is just a polynomial. For example, consider \(\left(D^{2}-D-6\right) y=10 x e^{3 x} ;\) multiplying both sides by \(e^{-3 x}\) and using \((\mathrm{c}),\) we get $$\begin{aligned} e^{-3 x}\left(D^{2}-D-6\right) y &=\left[(D+3)^{2}-(D+3)-6\right] y e^{-3 x} \\\ &=\left(D^{2}+5 D\right) y e^{-3 x}=10 x \end{aligned}$$ Show that a solution of \(\left(D^{2}+5 D\right) u=10 x\) is \(u=x^{2}-\frac{2}{5} x ;\) then \(y e^{-3 x}=x^{2}-\frac{2}{5} x\) or \(y=e^{3 x}\left(x^{2}-\frac{2}{5} x\right) .\) Use this method to solve Problems 23 to 26

Short answer

Use exponential shift and inverse exponential shift to transform and solve the differential equation. The final solution is \( y = e^{3 x}(x^{2} - \frac{2}{5} x) \).

Step-by-step solution

Show the Base Case for the Exponential Shift

Derivate the function to show the base cases. For the first derivative, apply the product rule: \( D\big(e^{ax} y\big) = e^{ax} \big(D + a\big) y \). Then apply the differentiation again for the second derivative: \( D^2\big(e^{ax} y\big) = e^{ax} \big(D + a\big)^2 y \).

Generalize for Any Power

Use mathematical induction to show that for any positive integer \(n\): \( D^n\big(e^{ax} y\big) = e^{ax} \big(D + a\big)^n y \). Assume it holds for a certain \( n \); then show it holds for \( n+1 \).

Show the Polynomial Form

If \( L(D) \) is any polynomial in the operator \( D \), we can generalize the solution from a linear combination to polynomials: \( L(D)\big(e^{ax} y\big) = e^{ax} L(D+a) y \).

Example Application (a)

Using the result from Step 3, solve specific operator equations. For instance: \( (D-1)^{3}\big(e^{x} y\big) = e^{x} D^{3} y \) and \( (D^{2}+D-6)\big(e^{-3 x} y\big) = e^{-3 x}(D^{2}-5 D)y \).

Derive the Inverse Exponential Shift

Replace \( D \) by \( D - a \) to find: \( e^{ax} P(D) y = P(D-a) e^{ax} y \)

Transform the Differential Equation

Use the inverse exponential shift method to transform the right-hand side of a given differential equation. For example, for \((D^{2} - D - 6)y = 10x e^{3x}\), multiply both sides by \(e^{-3x}\) and transform: \( e^{-3x}(D^{2} - D - 6)y = (D+3)^{2} - (D+3) - 6)y e^{-3x} = (D^{2}+5 D) y e^{-3 x} = 10 x \).

Solve the Transformed Equation

Solve the simpler polynomial equation: \( (D^{2} + 5 D) u = 10x \). Assume \( u(x) = x^{2} - \frac{2}{5} x \), then \( y e^{-3 x} = x^{2} - \frac{2}{5} x \).

Obtain the Final Solution

Multiply by \(e^{3x}\) to transform back and find the solution. \( y = e^{3 x}(x^{2} - \frac{2}{5} x)\).

Key concepts

Differential Operators

Differential operators, denoted commonly by \(D\), represent the process of differentiation. When you see \(D\) applied to a function, it means you need to take the derivative of that function.
For instance, if \(D\) acts on \(y\), then it simply gives us \(D(y) = y'\), the first derivative of \(y\). But differential operators can be much more powerful. They allow us to handle higher-order derivatives like \(D^2(y)\), which gives the second derivative, and so on.
One key feature of differential operators is their ability to be combined with other operations, such as multiplication and addition. This combination leads to polynomial expressions like \(D^2 + D - 6\), where each individual term represents a different level of differentiation. Such polynomials are crucial when solving differential equations because they simplify the handling of complex derivative terms. Understanding how differential operators work is the foundation to mastering more advanced topics like polynomial differential operators and exponential shifts.

Polynomial in Differential Operator

A polynomial in the differential operator \(D\) is an expression that involves \(D\) raised to various powers and combined with coefficients. For example, \(L(D) = D^2 + D - 6\) is a polynomial in \(D\).
This concept is important because it transforms the process of working with differential equations. Instead of manually differentiating multiple times, you can treat \(D\) as a single unit that follows polynomial rules.
To visualize this, consider the operation of applying \(L(D)\) to a function \(y\). We write it as \(L(D)y = (D^2 + D - 6)y\). Applying this operator involves taking the second derivative, first derivative, and a straightforward multiplication by -6. This operational approach saves time and helps avoid errors during manual differentiation.
Another crucial aspect is the relationship between polynomials in \(D\) and exponential shifts. When we shift the operator by a constant \(a\), the polynomial transforms as \(L(D+a)\). This shift property is powerful because it lets us handle complex expressions more elegantly by converting them into simpler forms.

Inverse Exponential Shift

The inverse exponential shift is an important technique for solving certain types of differential equations, especially those involving exponentials and polynomials. It involves shifting the differential operator by a negative constant, effectively 'undoing' the exponential shift.
To perform an inverse exponential shift, replace \(D\) with \(D-a\) in your polynomial operator, where \(a\) is the constant factor in the exponential. For example, converting \(P(D)\) into \(P(D-a)\).
Let's consider the exponential function \(e^{ax}\) and operator \(P(D)\). According to the inverse shift theorem: \[ e^{ax} P(D)y = P(D-a) e^{ax} y \]
This technique is a powerful tool to simplify differential equations. By multiplying both sides of a differential equation involving exponentials by the correct exponential factor, you can transform a complicated-looking problem into a simpler one that is easier to solve.
For instance, transforming the equation \((D^2 - D - 6)y = 10xe^{3x}\), you multiply by \(e^{-3x}\) to cancel out the exponential, transforming it to \(e^{-3x}(D^2 - D - 6)y = (D+3)^2 - (D+3) - 6)e^{-3x} y = (D^2 + 5D) y e^{-3 x} = 10x\). Solving this simpler polynomial equation then yields the general solution.