Solving Differential Equations: A Step-by-Step Guide

by TextBrain Team 53 views

Hey guys! Let's dive into the world of differential equations. Specifically, we're gonna tackle the problem: Consider the function y(x) given below: y(x)=2cx²+c², where c is a constant. Do the following: (a) Show that the given y(x) is a solution of the differential equation: (dy/dx)²+8x³dy/dx=16x²y (b) Find a particular solution. This might sound a bit intimidating at first, but trust me, we'll break it down step by step. We'll look at how to prove a function solves a differential equation and then find a specific solution. Get ready to flex those math muscles!

(a) Showing y(x) is a Solution

Alright, the first part of our mission is to prove that y(x) = 2cx² + c² actually solves the differential equation (dy/dx)² + 8x³(dy/dx) = 16x²y. This means that if we take our y(x), find its derivative, plug everything into the differential equation, and the equation holds true, then we've done it! Easy, right? Let's get started! The first thing we need is the derivative of y(x) with respect to x. Remember, c is a constant, so we'll treat it as such when we differentiate. Differentiating y(x) = 2cx² + c² gives us dy/dx = 4cx. Notice that the derivative of c² is zero because the derivative of a constant is always zero. Now that we have our dy/dx, we can plug both y(x) and dy/dx into the differential equation. So, let's substitute everything into the left-hand side (LHS) of the equation. We have (dy/dx)² + 8x³(dy/dx) = (4cx)² + 8x³(4cx) = 16c²x² + 32cx⁴. The right-hand side (RHS) of the equation is 16x²y = 16x²(2cx² + c²) = 32cx⁴ + 16c²x². Whoa! Check it out – the LHS equals the RHS! This means that the equation is true, so the function y(x) = 2cx² + c² is indeed a solution to the differential equation (dy/dx)² + 8x³(dy/dx) = 16x²y. See? Not so scary after all! We've successfully shown that our y(x) satisfies the given differential equation. That's part (a) done and dusted!

Now, let's recap what we've done. We started with our y(x), took its derivative, and then substituted both into the differential equation. The key takeaway here is that by doing so, we're essentially proving that the function we have is a valid solution for the differential equation. Understanding this process is key for more complex differential equations. Always start by taking the derivative, and then substitute. Once the left-hand side and the right-hand side are equal, then you are golden.

Key Points for Part (a):

  • Find the derivative: Calculate dy/dx from the given y(x).
  • Substitute: Plug both y(x) and dy/dx into the differential equation.
  • Simplify and Compare: Show that the LHS equals the RHS.

(b) Finding a Particular Solution

Now, on to the fun part: finding a specific solution! This means we need to find a particular value of c that satisfies some additional conditions. Without any additional conditions, y(x) = 2cx² + c² represents a general solution, where c can be any constant. A particular solution, however, requires us to find the value of c that makes the equation true given an additional condition. The problem did not provide us with this condition, therefore, we can assign a random value to c. For example, let's choose c = 1. We can then substitute this value of c back into our general solution, y(x) = 2cx² + c². We get y(x) = 2(1)x² + 1², which simplifies to y(x) = 2x² + 1. This is our particular solution. This y(x) not only satisfies the original differential equation but also meets the specific condition imposed by c = 1. Remember, there are infinitely many particular solutions depending on the value you assign to c. You could use c = 0, c = -2, or any other real number and generate a different, valid particular solution.

So, what does this all mean? Essentially, finding a particular solution is about pinpointing a specific solution out of the infinite possibilities described by the general solution. To do this, you need additional information (initial conditions, boundary values, etc.) or, in our case, just a c value. This extra information constrains the solution, giving you a unique outcome. Without the added condition, you are left to choose a c to get a value.

Key Points for Part (b):

  • Choose a Value for c: Since there are no additional conditions, select any value of c.
  • Substitute: Plug the value of c back into the general solution y(x).
  • Particular Solution: Simplify to get the particular solution. (i.e., y(x) = 2x² + 1)

Wrapping It Up

And that's a wrap, guys! We've successfully worked through both parts of the problem, demonstrating how to verify a solution to a differential equation and find a specific solution. We used the power of substitution to prove that our function y(x) solved the given differential equation. Then, we assigned a value to c to obtain a particular solution. The key takeaway here is the process: finding the derivative, substituting into the equation, and, in the case of a particular solution, utilizing additional conditions or values. Remember, differential equations are all about understanding how functions and their rates of change relate to each other. Keep practicing, and you'll become a differential equation whiz in no time! Now that you understand the basics, you can build on this knowledge and tackle more complex problems. Keep up the fantastic work! The more you practice, the more comfortable you'll get with these types of problems. Keep going!

Final Thoughts:

  • Review the concepts: Make sure you understand the difference between general and particular solutions.
  • Practice: Work through similar problems to solidify your understanding.
  • Ask Questions: Don't hesitate to seek help if you get stuck.

By following these steps, you'll be well on your way to mastering differential equations. Good luck and happy solving! Also, always remember to double-check your work, especially your derivatives and substitutions, to avoid making simple errors. This can save you a lot of time and frustration. Enjoy the process, guys!