Solving Linear PDEs: A Step-by-Step Guide

Hey guys! Let's dive into the world of solving linear partial differential equations (PDEs). I'll break down how to tackle some common types, covering homogeneous and non-homogeneous equations. We'll explore different methods and work through examples to make sure you've got a solid understanding. This guide is designed to be easy to follow, so grab a coffee, and let's get started!

a. Homogeneous PDE: $(D^6 + 2D^5 + 3D^4) y = 0$

Alright, first up, we've got a homogeneous PDE. This type of equation has a characteristic that, when it is set to zero, can be solved. The core idea here is to find a general solution that satisfies the equation. Remember, a homogeneous PDE has zero on the right-hand side. Let's crack into it!

To solve $(D^6 + 2D^5 + 3D^4) y = 0$, where $D$ represents the differential operator (d/dx), we begin by factoring the differential operator. This helps simplify the equation and identify the roots that will contribute to the solution. Factoring out the greatest common factor, we get $D^4(D^2 + 2D + 3) y = 0$. This shows that we have a repeated root from $D^4$ and a quadratic factor. Setting each factor to zero leads us to the roots of the characteristic equation. For $D^4 = 0$, we have a root of $0$ with multiplicity $4$. For the quadratic factor $D^2 + 2D + 3 = 0$, we can use the quadratic formula to find the roots. The roots are given by D = rac{-b rac{+}{} \sqrt{b^2 - 4ac}}{2a}. So, D = rac{-2 rac{+}{} \sqrt{2^2 - 4(1)(3)}}{2(1)} = rac{-2 rac{+}{} \sqrt{-8}}{2} = -1 rac{+}{} i\sqrt{2}. This tells us we have two complex conjugate roots. These roots will determine the form of our general solution.

Now, let's construct the general solution. For the root $0$ with multiplicity $4$, we get terms of the form $c_1 + c_2x + c_3x^2 + c_4x^3$, where $c_1, c_2, c_3,$ and $c_4$ are arbitrary constants. For the complex conjugate roots $-1 \pm i\sqrt{2}$, the solution will involve exponential and trigonometric functions. Specifically, the terms will be $e^{-x}(c_5 \cos(\sqrt{2}x) + c_6 \sin(\sqrt{2}x))$, where $c_5$ and $c_6$ are also arbitrary constants. Combining all these parts, the general solution to the homogeneous PDE is:

$y(x) = c_1 + c_2x + c_3x^2 + c_4x^3 + e^{-x}(c_5 \cos(\sqrt{2}x) + c_6 \sin(\sqrt{2}x))$.

This equation represents the general solution, encompassing all possible solutions to the original PDE. The constants $c_1$ through $c_6$ can be determined if we have initial or boundary conditions. See, not so bad, right?

b. Non-Homogeneous PDEs with Constant Coefficients

Alright, let's shift gears and look at non-homogeneous PDEs. These equations have a non-zero function on the right-hand side, which complicates things a bit. We'll solve these in different scenarios. The key here is to find both the complementary solution (from the homogeneous part) and a particular solution (to account for the non-homogeneous part). Let's break down some examples.

(i) $(D^2 + 4) y = 4 \cos 2x$

Let's get our hands dirty with the first example, $(D^2 + 4) y = 4 \cos 2x$. We have to use the method of undetermined coefficients or variation of parameters. The main objective is to deal with the non-homogeneous part of the equation. Here's how we'll approach it:

First, let's find the complementary solution. We solve the homogeneous equation $(D^2 + 4) y = 0$. The characteristic equation is $r^2 + 4 = 0$, so $r = \pm 2i$. Thus, the complementary solution is $y_c(x) = c_1 \cos(2x) + c_2 \sin(2x)$. Note how these are solutions to the homogenous part of the differential equation.

Next, we'll find a particular solution. Because the right-hand side is $4 \cos 2x$, which is part of the complementary solution, we have to modify our guess. Usually, we'd guess $A \cos(2x) + B \sin(2x)$, but since it overlaps with $y_c$, we will use $y_p(x) = x(A \cos(2x) + B \sin(2x))$. We take the derivatives of $y_p(x)$: $y'_p(x) = A \cos(2x) + B \sin(2x) + 2x(-A \sin(2x) + B \cos(2x))$, and $y''_p(x) = 4B \cos(2x) - 4A \sin(2x) - 4A \cos(2x) - 4B \sin(2x)$. Substituting these into the original equation, we get $(D^2 + 4) y_p(x) = 4 \cos 2x$. This simplifies to $-4A \cos(2x) - 4B \sin(2x) + 4(A \cos(2x) + B \sin(2x)) = 4 \cos(2x)$. We see that the sine terms cancel, and we have $-4A = 4$, so $A = -1$. Then we solve for B, and $B = 0$. So our particular solution is now $y_p(x) = -x \cos(2x)$.

Finally, we combine the complementary and particular solutions to get the general solution: $y(x) = y_c(x) + y_p(x) = c_1 \cos(2x) + c_2 \sin(2x) - x \cos(2x)$.

(ii) $(D^3 - 2D^2 + 10D) y = 4 e^{2x}$

Time for the next one, $(D^3 - 2D^2 + 10D) y = 4 e^{2x}$. Again, we start with the complementary solution, find a particular solution, then combine the solutions.

For the complementary solution, we solve the homogeneous equation $(D^3 - 2D^2 + 10D) y = 0$. This is where we factor the operator. $D(D^2 - 2D + 10) y = 0$. The characteristic equation is $r(r^2 - 2r + 10) = 0$. Thus, the roots are $r = 0$ and r = rac{2 \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)} = 1 \pm 3i. The complementary solution is $y_c(x) = c_1 + e^x(c_2 \cos(3x) + c_3 \sin(3x))$.

Next, find the particular solution. Here, the right-hand side is $4e^{2x}$, so we guess a solution of the form $y_p(x) = Ae^{2x}$. Substitute this into the original equation to solve for A. $D y_p = 2Ae^{2x}$, $D^2 y_p = 4Ae^{2x}$, and $D^3 y_p = 8Ae^{2x}$. Plugging them into the original equation: $(8A - 2(4A) + 10(2A))e^{2x} = 4e^{2x}$. This simplifies to $(8A - 8A + 20A)e^{2x} = 4e^{2x}$, so $20A = 4$, giving $A = 1/5$. The particular solution is y_p(x) = rac{1}{5}e^{2x}.

Combining both solutions, the general solution is $y(x) = c_1 + e^x(c_2 \cos(3x) + c_3 \sin(3x)) + \frac{1}{5}e^{2x}$.

(iii) $(D^2 - 9) y = x^2 e^{2x}$

Here we go, the last one! For $(D^2 - 9) y = x^2 e^{2x}$, let's find a solution.

For the complementary solution, we solve $(D^2 - 9) y = 0$, and the characteristic equation is $r^2 - 9 = 0$. Thus, $r = \pm 3$. The complementary solution is $y_c(x) = c_1e^{3x} + c_2e^{-3x}$.

Next, we find the particular solution. Given the form $x^2 e^{2x}$, we use the method of undetermined coefficients. We guess $y_p(x) = (Ax^2 + Bx + C)e^{2x}$. Calculating the derivatives, we get $y'_p(x) = (2Ax + B)e^{2x} + 2(Ax^2 + Bx + C)e^{2x}$, and $y''_p(x) = (2A + 4Ax + 2B)e^{2x} + 2(2Ax + B)e^{2x} + 4(Ax^2 + Bx + C)e^{2x}$. After substituting these into the original equation and simplifying and equating coefficients, it's found that A = rac{1}{13}, $B = -\frac{4}{169}$, and $C = \frac{76}{2197}$. The particular solution is $y_p(x) = (\frac{1}{13}x^2 - \frac{4}{169}x + \frac{76}{2197})e^{2x}$.

Finally, the general solution is $y(x) = c_1e^{3x} + c_2e^{-3x} + (\frac{1}{13}x^2 - \frac{4}{169}x + \frac{76}{2197})e^{2x}$.

c. PDEs with…

This section has been purposefully left incomplete to maintain brevity and clarity in the provided examples. Additional PDE scenarios would be introduced here, building upon the principles explained above. This segment would continue with more complex PDE problems. These scenarios would vary based on the problem's unique requirements, making it a flexible resource for various PDE types.

In summary, solving linear PDEs involves identifying the equation type, determining the complementary and particular solutions, and combining them. Remember to check your work, and don't be afraid to seek additional resources if you need them. Good luck, and happy solving!"