Problem

Solve the following initial-value problems by usi…

03:23

Problem 249 Hard Difficulty

Solve the following initial-value problems by using integrating factors.
$$
\sqrt{x} y^{\prime}=y+2 x, y(0)=1
$$

Answer

$y(x)=-2 x-2 x^{1 / 2}+2 e^{2 x^{1 / 2}}-1$

Related Courses

Calculus 2 / BC

Calculus

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

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Video Transcript

Section four without five problem to 49. We're dealing with first order linear ordinary differential equations. The key to this equation. First, to get it in standard form, why prime plus a function of X times Why is equal to another function off X? So to place this one in standard form that's going to be y prime minus one over root X Why is equal to two X divided by the square root of acts? So that's just gonna be too square root of X. So the integrating factor is going to be e to the minus one over root X d x. And so this is going to be e tu minus two root X So that's my integrating factor. So this equation turns into why e to the minus two root X prime is equal to to root x you to the minus two root x. Now, then the key becomes integrating both sides of this equation. So to integrate both sides of this equation on the left, I get why e tu minus two root X is equal to know when I integrate on the right side, I'm gonna end up with two x plus two x plus one times E to the minus two root X plus a constant of integration. Now I will solve for wine by dividing by E to the minus two root x and that leave me with minus two x plus to root X plus one. So all of this plus see you to the to root X. You have the boundary condition that y zero is one. So that would give me that one is equal to minus zero plus zero plus one plus c and then e to the zero is just going to be one there. So this tells me that one is equal to minus one plus C. Therefore, C is equal to two. So this final answer is why is equal to minus two X plus to root X plus one and then what states use it with the two plus two e to the to root X, and that is our final answer.

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