Solve the following initial value problem:dydx=2x−5\frac{dy}{dx}=2x-5dxdy=2x−5; y(0)=4y\left(0\right)=4y(0)=4

y ( x ) = x 2 − 5 x + 4 y\left(x\right)=x^2-5x+4 y ( x ) = x 2 − 5 x + 4

Solve the following initial value problem: d y d x = 2 x − 5 \frac{dy}{dx}=2x-5 d x d y = 2 x − 5 ; y ( 0 ) = 4 y\left(0\right)=4 y ( 0 ) = 4

this problem, we're asked to solve the following IVP or initial value problem. So notice we have this derivative is equal to 2x minus 5 given to us right here. We also have this initial condition, and we're asked to solve the initial value problem. So let's go ahead and get started. Now my first step is going to be to find the function y of x. Since I can see we're given the derivative of y with respect to x, I'm gonna work backwards here with my derivatives to figure out what y of x is. And I can do that by integrating the right side of this equation. So what that's going to be is the integral of 2x minus 5 all with respect to x. So integrating this right side here, well, I can use the rules for integrals that I know. I can integrate the 2x, and then I can integrate the 5 separately because we have a sum or difference rule that we're applying right about here. I can also take this constant of 2 and bring it outside the integral. So we're going to have 2 times the integral of x with respect to x. You're going to have minus the integral of 5 with respect to x. I'll leave the 5 inside the integral because this constant is by itself. Now what I can do is use the power rule, where we add 1 to the exponent and then divide by that new exponent. And that's going to allow me to evaluate these integrals. So doing this, I'm going to get 2 times, and I'm going to have x squared over 2. That's doing this integral right here. Then we'll have minus, and then the integral of 5 is just going to be 5 x, then we'll have plus the constant of integration c. Now I'll go ahead and I'll cancel these twos right here, giving me that my function y of x is equal to x squared minus 5x plus c. Now this is the general solution right here. And now that I find the general solution, what I want to do is I want to find the particular solution given this condition that I have. So what I can do is take 0 and replace every place that I see x with 0 in this function. So doing this, we're going to move on to this portion right here where we're going to have that y of 0 is equal to, in this case, is gonna be 0 squared minus 5 times 0 plus the constant c. Now notice every place that we see 0 is just going to go to 0. So we're going to have 0 minus 0 plus c, which means that y of 0 is simply going to be equal to c. And I can see here that y of 0 is equal to 4, meaning that c is equal to 4. So 4 is my solution for c. So going back up to this general solution, I found I just take that c and replace it with 4. So this right here would be the solution for my function, and that is you can solve this initial value problem. Hope you found this video helpful.

Solve the following initial value problem: d y d x = 2 x − 5 \frac{dy}{dx}=2x-5 d x d y = 2 x − 5 ; y ( 0 ) = 4 y\left(0\right)=4 y ( 0 ) = 4

y ( x ) = x 2 − 5 x + 4 y\left(x\right)=x^2-5x+4 y ( x ) = x 2 − 5 x + 4

y ( x ) = 2 x 2 − 5 x y\left(x\right)=2x^2-5x y ( x ) = 2 x 2 − 5 x

y ( x ) = x 2 − 5 x y\left(x\right)=x^2-5x y ( x ) = x 2 − 5 x

y ( x ) = 2 x 2 − 5 x + 4 y\left(x\right)=2x^2-5x+4 y ( x ) = 2 x 2 − 5 x + 4

7. Antiderivatives & Indefinite Integrals

Initial Value Problems

Calculus

7. Antiderivatives & Indefinite Integrals

Initial Value Problems

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Multiple Choice

Solve the following initial value problem:
$\frac{dy}{dx}=2x-5$ ; $y\left(0\right)=4$

$y\left(x\right)=2x^2-5x$

$y\left(x\right)=x^2-5x$

$y\left(x\right)=2x^2-5x+4$

$y\left(x\right)=x^2-5x+4$

Verified step by step guidance

Start by identifying the given differential equation: $ \frac{dy}{dx} = 2x - 5 $. This is a first-order differential equation.

To solve this differential equation, integrate both sides with respect to $ x $. The left side becomes $ y $ and the right side requires integration: $ \int (2x - 5) \, dx $.

Perform the integration: $ \int 2x \, dx = x^2 $ and $ \int -5 \, dx = -5x $. Combine these results to get $ y = x^2 - 5x + C $, where $ C $ is the constant of integration.

Use the initial condition $ y(0) = 4 $ to find the value of $ C $. Substitute $ x = 0 $ and $ y = 4 $ into the equation: $ 4 = 0^2 - 5(0) + C $, which simplifies to $ C = 4 $.

Substitute $ C = 4 $ back into the equation to get the particular solution: $ y(x) = x^2 - 5x + 4 $.

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