Solving Differential Equations: A Step-by-Step Guide

Hey math enthusiasts! Let's dive into the fascinating world of differential equations and figure out how to solve some classic problems. We'll break down the solutions to three specific differential equations, making sure everything is clear and easy to follow. Get ready to flex those math muscles!

Differential Equation 1: 2xyy' = y² – x²

Alright, first up, let's tackle the differential equation: 2xyy' = y² – x². This one might look a little intimidating at first, but trust me, we can break it down. Our goal is to find a function y(x) that satisfies this equation. The key here is recognizing that this equation is a homogeneous differential equation.

What does homogeneous even mean? Well, it means that if you replace x with tx and y with ty, the equation remains the same. Let's see how that works. Substitute tx for x and ty for y into the equation: 2(tx)(ty)y' = (ty)² – (tx)². Simplify this to get 2t²xyy' = t²y² – t²x². Notice that we can divide both sides by t² and end up with our original equation 2xyy' = y² – x². This confirms it's homogeneous.

Since it's homogeneous, we can use a substitution to simplify things. Let's use the substitution v = y/x. This implies y = vx. Now, we need to find y' to substitute into our original equation. Differentiating y = vx with respect to x using the product rule gives us y' = v + xv'.

Now, let's substitute y = vx and y' = v + xv' into our original equation. We get 2x(vx)(v + xv') = (vx)² – x². Simplify this to get 2x²v(v + xv') = x²v² – x². Divide everything by x² (assuming x ≠ 0) and we have 2v(v + xv') = v² – 1. This simplifies further to 2v² + 2xvv' = v² – 1. Now, let's rearrange to isolate the terms involving v and x: 2xvv' = v² – 1 – 2v². That becomes 2xvv' = -v² – 1. We're getting closer!

Next, isolate v' and separate the variables. Rearranging gives us v' = (-v² - 1) / (2xv). Separate the variables by multiplying both sides by dx and dividing both sides by (-v² - 1) and by x. We get (2v / (-v² - 1))dv = (1 / x)dx. Now we integrate both sides: integral of (2v / (-v² - 1))dv = integral of (1 / x)dx. Solving the integrals we get -ln(|v² + 1|) = ln(|x|) + C, where C is the constant of integration.

We can simplify this by exponentiating both sides. We get |v² + 1| = e^(-ln(|x|) - C). Simplify to get v² + 1 = ±e^(-C) / x. Let's replace ±e^(-C) with another constant K, which gives us v² + 1 = K / x. Then, solving for v, we get v = ±√((K / x) - 1). Finally, we substitute back y/x for v, obtaining y/x = ±√((K / x) - 1). Therefore, the solution for y is y = ±x√((K / x) - 1). Boom! We've found the solution. Always remember to check your work and ensure your solution satisfies the original equation. We did it, guys!

Differential Equation 2: xy' = x + y

Now, let's tackle the second differential equation: xy' = x + y. This equation is a bit simpler than the first one. We can approach this one using a slightly different method. Here, we can rearrange the equation to resemble a linear first-order differential equation. Let's divide both sides by x (assuming x ≠ 0) to get y' = 1 + y/x. Then rearrange to y' - (1/x)y = 1. This is a first-order linear differential equation in the form y' + P(x)y = Q(x).

To solve this, we need to find an integrating factor, which is e^(integral of P(x)dx). In this case, P(x) = -1/x. So, the integral of P(x)dx = integral of (-1/x)dx = -ln(|x|). The integrating factor is e^(-ln(|x|)). Using the properties of logarithms and exponents we simplify this to 1/|x|. For the sake of this problem, let's assume x > 0, so our integrating factor becomes 1/x.

Next, multiply both sides of the rearranged equation (y' - (1/x)y = 1) by the integrating factor (1/x). This gives us (1/x)y' - (1/x² )y = 1/x. The left-hand side can be rewritten as the derivative of the product of y and the integrating factor: ((1/x)y)' = 1/x. Now, integrate both sides with respect to x. So, integral of ((1/x)y)'dx = integral of (1/x)dx. This gives us (1/x)y = ln(|x|) + C.

Finally, to solve for y, we multiply both sides by x: y = xln(|x|) + Cx. And that, my friends, is our solution! Therefore the solution to the differential equation xy' = x + y is y = xln(|x|) + Cx where C is the constant of integration. Another one bites the dust!

General Approach

• Rearrange: Isolate y' and rearrange to a standard form. This might involve dividing by x or other terms.
• Identify: Recognize the equation type. (e.g., linear).
• Integrate: Find an integrating factor and apply it to the equation. Then, integrate both sides to find y. Don’t forget the constant of integration!

Differential Equation 3: xy' = y² + y

Alright, let's wrap things up with the differential equation: xy' = y² + y. This equation looks a little different. We can see that this is a separable differential equation because we can separate the variables x and y. To solve this, we'll aim to get all the y terms on one side and all the x terms on the other. First, rewrite the equation as y' = (y² + y) / x. Then, separate variables to get (1 / (y² + y))dy = (1 / x)dx.

Before we integrate, we need to simplify the left side. We can use partial fraction decomposition to break down 1 / (y² + y). Factor the denominator to get y(y + 1), so we need to rewrite 1 / (y(y + 1)) in the form of A/y + B/(y + 1). Multiplying both sides by y(y + 1), we get 1 = A(y + 1) + By. If we let y = 0, we get A = 1. If we let y = -1, we get B = -1. So, we can rewrite the left side as (1/y) - (1/(y + 1)). Now we can integrate both sides: integral of ((1/y) - (1/(y + 1)))dy = integral of (1/x)dx.

Integrating both sides, we get ln(|y|) - ln(|y + 1|) = ln(|x|) + C. We can simplify the left side using the properties of logarithms. We get ln(|y / (y + 1)|) = ln(|x|) + C. Now, let's exponentiate both sides to get rid of the logarithms: |y / (y + 1)| = e^(ln(|x|) + C). This simplifies to |y / (y + 1)| = e^C * |x|. We can replace e^C with another constant, let's say K, so we have |y / (y + 1)| = K|x|.

Next, remove the absolute value signs by writing y / (y + 1) = Kx. Now we need to solve for y. Multiply both sides by (y + 1) to get y = Kxy + Kx. Rearrange the terms to isolate y: y - Kxy = Kx. Factor out y: y(1 - Kx) = Kx. Finally, divide by (1 - Kx) to solve for y. The solution is y = Kx / (1 - Kx), where K is the constant of integration. Awesome! We've successfully solved all three differential equations. Pat yourselves on the back, guys and gals! Remember, practice makes perfect. Keep working through these problems and you'll become a differential equation master in no time! Keep practicing, and you'll be solving these in your sleep!