Solving Linear Differential Equations: A Step-by-Step Guide

Hey guys! Today, we're diving deep into the fascinating world of differential equations, specifically focusing on how to solve a linear differential equation. We'll break down the process step-by-step, making it super easy to understand, even if you're just starting out. Our example equation is xy' + 3y = x². So, grab your pencils, and let's get started!

Understanding Linear Differential Equations

Before we jump into solving, let's understand what we're dealing with. A linear differential equation is basically an equation that involves an unknown function and its derivatives, where the function and its derivatives appear linearly. Think of it like a recipe where you're trying to find the secret ingredient (the function) based on how it changes (its derivatives). These equations pop up all over the place in science and engineering, from modeling population growth to analyzing electrical circuits. The general form of a first-order linear differential equation looks like this:

dy/dx + P(x)y = Q(x)

Where:

• dy/dx is the first derivative of y with respect to x.
• P(x) is a function of x.
• y is the unknown function we're trying to find.
• Q(x) is another function of x.

Our equation, xy' + 3y = x², fits this general form, which is excellent news! Recognizing this form is the first crucial step in solving these types of problems. It tells us which tools and techniques we can use. Now, let's move on to prepping our equation for the main event: finding the solution!

Step 1: Standard Form - Getting Ready to Solve

The first thing we need to do is rewrite our equation, xy' + 3y = x², into the standard form mentioned earlier: dy/dx + P(x)y = Q(x). This form makes it much easier to identify the key components we need for our solution. To get there, we need to get rid of the x that's hanging out in front of the y' term (which is the same as dy/dx). We can do this by dividing every single term in the equation by x. It's crucial to remember to divide everything to keep the equation balanced and mathematically sound. Think of it like equally distributing ingredients in a cake recipe – you want the right proportion in every slice!

So, let's do it:

(xy'/x) + (3y/x) = (x²/x)

This simplifies beautifully to:

y' + (3/x)y = x

Ta-da! We've successfully transformed our equation into the standard form. Now, we can clearly see that:

• P(x) = 3/x
• Q(x) = x

Identifying P(x) and Q(x) is super important because these guys are the stars of the next step: finding the integrating factor. This factor will act like a magic key, unlocking the solution to our differential equation. We're making great progress – keep going!

Step 2: The Integrating Factor - Our Magic Key

Okay, now for the fun part: finding the integrating factor. This is a special function that helps us solve the differential equation. It's like a secret ingredient that makes everything work perfectly! The integrating factor, often denoted by μ(x) (that's the Greek letter 'mu'), is calculated using the following formula:

μ(x) = e^(∫P(x) dx)

Don't let the formula scare you! It's not as complicated as it looks. Remember that P(x) is the function we identified in the previous step (the one multiplying the y term in the standard form). The integral symbol (∫) just means we need to find the antiderivative of P(x), and e is the base of the natural logarithm (a special number approximately equal to 2.718). So, let's plug in our P(x) = 3/x:

μ(x) = e^(∫(3/x) dx)

Now we need to find the integral of 3/x. Remember your calculus? The integral of 1/x is ln|x| (the natural logarithm of the absolute value of x). Since we have a 3 multiplying the 1/x, we can pull that constant out of the integral:

∫(3/x) dx = 3∫(1/x) dx = 3ln|x|

So, our integrating factor becomes:

μ(x) = e^(3ln|x|)

We can simplify this even further using a property of logarithms: aln(b) = ln(b^a). Applying this, we get:

μ(x) = e^(ln|x³|)

And finally, since the exponential function and the natural logarithm are inverse operations, they cancel each other out (e^(ln(a)) = a):

μ(x) = |x³|

For simplicity, and because we are generally interested in intervals where x is positive, we can drop the absolute value signs and say:

μ(x) = x³

Awesome! We've found our integrating factor. This x³ is the key to unlocking the solution. Get ready for the next step, where we'll put this magic key to work!

Step 3: Multiplying Through - Unleashing the Integrating Factor

Now that we've found our integrating factor, μ(x) = x³, it's time to put it to work. This step is crucial, as it transforms the left-hand side of our differential equation into a form that we can easily integrate. Remember our standard form equation:

y' + (3/x)y = x

We're going to multiply every single term in this equation by our integrating factor, x³. Just like in step one, multiplying every term ensures we maintain the equality. Think of it like amplifying a signal – you need to amplify the entire signal, not just parts of it, to get the complete picture.

So, let's multiply:

x³(y') + x³((3/x)y) = x³(x)

This simplifies to:

x³y' + 3x²y = x⁴

Now, here's the magic part. The left-hand side of this equation is now the derivative of a product! It's the derivative of (y * μ(x)), which in our case is (y * x³). This is the whole reason we went through the trouble of finding the integrating factor. It cleverly transforms the equation into something we can directly integrate.

To see why this is true, remember the product rule for differentiation: d/dx(uv) = u'v + uv'. If we let u = y and v = x³, then:

d/dx(yx³) = y'x³ + y(3x²) = x³y' + 3x²y

And that's exactly what we have on the left-hand side of our equation! So, we can rewrite our equation as:

d/dx(yx³) = x⁴

This is a huge step forward. We've gone from a differential equation that looks tricky to solve to one where we can isolate the derivative of a known product. Next up, we'll integrate both sides to get rid of that derivative and finally find our solution!

Step 4: Integration - Finding the General Solution

We've reached the point where we can finally integrate both sides of our equation. Remember, we're trying to get rid of the derivative, and integration is the inverse operation of differentiation. Our equation from the previous step was:

d/dx(yx³) = x⁴

To get rid of the d/dx on the left, we integrate both sides with respect to x:

∫[d/dx(yx³)] dx = ∫x⁴ dx

The integral of a derivative just gives us the original function, so the left side becomes:

yx³ = ∫x⁴ dx

Now, we need to evaluate the integral on the right-hand side. Recall the power rule for integration: ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. Applying this rule, we get:

∫x⁴ dx = (x⁵)/5 + C

So, our equation now looks like this:

yx³ = (x⁵)/5 + C

This is the general solution to our differential equation. It's called the general solution because it includes an arbitrary constant, C. This constant represents a family of solutions, each corresponding to a different value of C. To find a specific solution (a particular solution), we would need an initial condition (a value of y at a specific x). But for now, we've got the general solution, which is a huge accomplishment!

Step 5: Solving for y - Isolating Our Function

We're almost there! We have the general solution, but it's not quite in its most user-friendly form. We want to isolate y on one side of the equation to clearly see the function we've found. Our equation currently looks like this:

yx³ = (x⁵)/5 + C

To get y by itself, we simply need to divide both sides of the equation by x³:

(yx³)/x³ = [(x⁵)/5 + C] / x³

This simplifies to:

y = (x⁵)/(5x³) + C/x³

We can simplify the first term further by canceling out x³ from the numerator and denominator:

y = x²/5 + C/x³

And there you have it! We've successfully solved the linear differential equation. This is our final, simplified general solution:

y = x²/5 + Cx⁻³

This equation tells us the family of functions that satisfy our original differential equation, xy' + 3y = x². Remember that C is an arbitrary constant, so there are infinitely many solutions, each corresponding to a different value of C. If we were given an initial condition, we could plug it in to solve for C and find the specific solution that meets that condition.

Conclusion

Great job, guys! We've successfully navigated the process of solving a linear differential equation. We took the equation xy' + 3y = x², transformed it into standard form, found the integrating factor, multiplied through, integrated, and finally solved for y. It might seem like a lot of steps, but each step is logical and builds upon the previous one. The key is to break it down and understand the reasoning behind each action.

Remember, practice makes perfect! The more you work through these types of problems, the more comfortable you'll become with the process. Differential equations might seem daunting at first, but with a little perseverance, you can conquer them. Keep practicing, and you'll be solving differential equations like a pro in no time! Good luck, and happy solving!