Find the general solution to the differential equation.dydx=yx\frac{dy}{dx}=y\sqrt{x}dxdy=yx

y = C e 2 3 x 3 2 y=Ce^{\frac23x^{\frac32}} y = C e 3 2 x 2 3

Find the general solution to the differential equation. d y d x = y x \frac{dy}{dx}=y\sqrt{x} d x d y = y x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z">

this problem, we're asked to find the general solution to the differential equation dy dx equals y times the square root of x. Now looking at this equation, we can clearly see that this is separable as it's already explicitly written in the form that dy dx is equal to some function of x multiplying some function of y. So we wanna go ahead and get started with our step steps here where the first thing we wanna do is separate our variables. So if d y d x is equal to y times the square root of x, we want to get that function of y and d y together on one side, our function of x and d x together on the other. So the first thing I'm going to do here is divide both sides by y. Remember, that's the same thing as multiplying by one over y. So now I have moved that over, and I can go ahead and move that d x over in the same step as I multiply both sides by d x in order to cancel that out on the left. So now I have one over y dy equals the square root of x times dx. So with those variables separated, I can move on to my next step, which is integrating both sides. So integrating each side here, the integral of one over y d y is going to be the natural log of the absolute value of y, and that's then equal to the integral of the square root of x d x. We can apply our power rule here remember the square root of x is the same thing as x to the power of one half. So So this is gonna give us here two thirds x to the power of three halves, and then plus our constant of integration c. So we finished that step two. And now since we're asked to find the general solution, we can go ahead and move on to step four, skipping over that step three since we're not given an initial condition with which we could find c. So now we're just solving for y. So I wanna get this y all by itself, and I can do that by taking e and raising it to each side. So that e and natural log will cancel out, making this the absolute value of y, and that's then equal to e to the power of two thirds, x to the power of three halves, and then plus c. Now you'll see a couple of things happen here that may get skipped whenever you watch other solutions, and that's because oftentimes, tutors will do this all in one step. But I'm gonna go ahead and break this down for you. There are a couple of things that we wanna change. We wanna get this c in front of that e by rewriting this using exponential rules. So I can rewrite this as e to the power of two thirds, x to the power of three halves times e to the power of c. Now this all by itself is just another constant, so we could effectively just call this c. Now you may see people refer to this as a different constant, say c one, but it doesn't matter because in the end, it all just ends up being a constant. So I'm gonna go ahead and keep that as c. So now I can rewrite this as c times e to the power of two thirds, x to the power of three halves. And this is equal to the absolute value of y. Now you may be wondering here how we get rid of these absolute value signs and why oftentimes they seem to just get ignored. And this is because in the same step, we can go ahead and rewrite this right hand side as a plus or minus c times e to the power of two thirds x to the power of three halves. But oftentimes we're just gonna absorb that plus or minus altogether in that c because no matter if it's positive or negative it's still a constant. So we can really just write this as y equals c times e to the power of two thirds x to the power of three halves, and this is our final answer. Now everything that I did in these final couple of steps is something that you may often see get skipped, but just know that that's really what's happening here as we get to our final answer. Let us know if you have any questions, and I'll see you in the next one.

Find the general solution to the differential equation. d y d x = y x \frac{dy}{dx}=y\sqrt{x} d x d y = y x <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

y = C e 2 3 x 3 2 y=Ce^{\frac23x^{\frac32}} y = C e 3 2 x 2 3

y = 2 3 x 3 2 + C y=\frac23x^{\frac32}+C y = 3 2 x 2 3 + C

y = 3 2 x 3 2 + C y=\frac32x^{\frac32}+C y = 2 3 x 2 3 + C

y = C e 2 3 x 2 3 y=Ce^{\frac23x^{\frac23}} y = C e 3 2 x 3 2

13: Intro to Differential Equations

Separable Differential Equations

Business Calculus

13: Intro to Differential Equations

Separable Differential Equations

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Multiple Choice

Find the general solution to the differential equation.
$\frac{dy}{dx}=y\sqrt{x}$

$y=\frac23x^{\frac32}+C$

$y=\frac32x^{\frac32}+C$

$y=Ce^{\frac23x^{\frac32}}$

$y=Ce^{\frac23x^{\frac23}}$

Verified step by step guidance

Step 1: Start by identifying the type of differential equation. The given equation is a first-order differential equation of the form $ \frac{dy}{dx} = y \sqrt{x} $. This suggests that it may be separable, meaning we can rewrite it to separate the variables $ y $ and $ x $.

Step 2: Rewrite the equation to separate the variables. Divide both sides by $ y $ and multiply by $ dx $ to get $ \frac{1}{y} dy = \sqrt{x} dx $. This allows us to integrate each side independently.

Step 3: Integrate both sides. The left-hand side becomes $ \int \frac{1}{y} dy = \ln|y| $, and the right-hand side becomes $ \int \sqrt{x} dx = \frac{2}{3}x^{\frac{3}{2}} + C_1 $, where $ C_1 $ is the constant of integration.

Step 4: Combine the results of the integration. We now have $ \ln|y| = \frac{2}{3}x^{\frac{3}{2}} + C_1 $. Exponentiate both sides to solve for $ y $, yielding $ y = e^{\frac{2}{3}x^{\frac{3}{2}} + C_1} $. Since $ e^{C_1} $ is just a constant, we can write $ y = Ce^{\frac{2}{3}x^{\frac{3}{2}}} $, where $ C $ is a constant.

Step 5: Verify the solution. Substitute $ y = Ce^{\frac{2}{3}x^{\frac{3}{2}}} $ back into the original differential equation $ \frac{dy}{dx} = y \sqrt{x} $ to confirm that it satisfies the equation. This ensures the solution is correct.

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