Find the general solution to the differential equation dydx=−2x+5x2\frac{dy}{dx}=-2x+5x^2dxdy=−2x+5x2.

y = − x 2 + 5 3 x 3 + C y=-x^2+\frac53x^3+C y = − x 2 + 3 5 x 3 + C

Find the general solution to the differential equation d y d x = − 2 x + 5 x 2 \frac{dy}{dx}=-2x+5x^2 d x d y = − 2 x + 5 x 2 .

we're asked to find the particular solution to the differential equation y prime equals two e to the t plus four t, given the initial condition that y of zero is equal to one. So in order to find our particular solution, we first need to just find our general solution. So if y prime is equal to two e to the power of t plus four t, we can go ahead and integrate both sides in order to just get y. So y is then going to be equal to the integral of two e to the t plus four t and then of course we're integrating with respect to t. So finding our integral here that first term that's just gonna stay the same two e to the t because it's just an exponential function And then for this next term here, four t, we wanna go ahead and apply our power rule adding one to that exponent, dividing by that new exponent. That's going to give us here two t squared. Then because this is an indefinite integral, we wanna go ahead and add on our constant of integration c. So this gives us here our general solution, but that's not what we're asked to find here. We need to take this a step further and find our particular solution based on our initial condition that y of zero is equal to one. So we want to go ahead and plug zero in for x and one in for y into our general solution here so that we can solve for c. So plugging that one in for y, this gives me one equals two e to the power of zero. Since here we are actually working with t instead of x, we're going to be plugging into this zero for t just because those are the variables we're using in this equation specifically. So two e to the zero plus two times zero squared plus C. Now anything times zero is zero, so this entire term is going to go away. And then e to the zero, that's just going to give us one. So I have one equals two plus C. Then if I subtract two on both sides to get that C all by itself, one minus two gives me negative one. So negative one is then equal to C. So now I can get my particular solution by plugging that C back into my general solution here. So this gives me a y equals two e to the power of t plus two t squared minus one. And this is my particular solution to the given differential equation based on that initial condition that y of zero is equal to one. Feel free to 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 = − 2 x + 5 x 2 \frac{dy}{dx}=-2x+5x^2 d x d y = − 2 x + 5 x 2 .

y = − x 2 + 5 3 x 3 + C y=-x^2+\frac53x^3+C y = − x 2 + 3 5 x 3 + C

y = − x 2 + 10 x + C y=-x^2+10x+C

y = − x + 5 3 x 2 + C y=-x^{}+\frac53x^2+C y = − x + 3 5 x 2 + C

13: Intro to Differential Equations

Basics of Differential Equations

Business Calculus

13: Intro to Differential Equations

Basics of Differential Equations

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Multiple Choice

Find the general solution to the differential equation $\frac{dy}{dx}=-2x+5x^2$ .

$y=-x^2+\frac53x^3+C$

$y = - x^{2} + 10 x + C$

$y = - x + 5 x + C$

$y=-x^{}+\frac53x^2+C$

Verified step by step guidance

Step 1: Start with the given differential equation $ \frac{dy}{dx} = -2x + 5x^2 $. This is a first-order differential equation, and we aim to find its general solution by integrating both sides.

Step 2: Rewrite the equation to isolate $ dy $ and $ dx $: $ dy = (-2x + 5x^2) dx $. This allows us to integrate with respect to $ x $.

Step 3: Break the right-hand side into separate terms for easier integration: $ \int dy = \int (-2x) dx + \int (5x^2) dx $.

Step 4: Integrate each term separately. For $ \int (-2x) dx $, use the power rule $ \int x^n dx = \frac{x^{n+1}}{n+1} $, resulting in $ -x^2 $. For $ \int (5x^2) dx $, apply the same rule to get $ \frac{5}{3}x^3 $.

Step 5: Combine the results and add the constant of integration $ C $: $ y = -x^2 + \frac{5}{3}x^3 + C $. This is the general solution to the differential equation.

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