# Wondering if these two First Linear Order IVPs are correct

• garr6120
In summary, the equation for number 1 is incorrect, and for number 2, the integrating factor has not been found.
garr6120

## Homework Statement

I am having trouble proving if the equation i have found for number 1 is correct. I have posted my solution to get back to the main problem in the first photo below.

For number 2 I am having trouble isolating for 1 y(x). Did i do the integration and setup properly?

## Homework Equations

Question 1:

given y' + ycotanx+2cosx (1) and y(π/2)=0 find the IVP.

find the integrating factor from equation 1.
μ(x)=e∫cotanxdx=sinx (2)

multiply equation 2 by 1.
y'sinx+ysinxcotanx=2cosxsinx
(sinxy)'=2cosxsinx (3)

integrate equation 3.
sinxy=∫2cosxsinx=sin2x+c (4)

Isolate y in 4.
y=sinx+c(csc2x)

plug in the initial value to find c which is found to be -1.
y=sinx-csc2x (5)

proving that this is a solution to the differential:
take the derivative of equation 5.
y'=cosx+2csc2xcotanx

plug y and y' into differential equation 1.
y'+ycotanx=cosx+2csc2xcotanx+cosx-cotanxcsc2x
2cosx+3csc2xcotanx

I cannot get an answer for 2cosx. I did everything right.

For Question 2:

y'+ytanx=y2 (1) for y(0)=1/2

find an integration factor.
μ(x)=e∫tanxdx=1/cosx {2}

multiply equation 1 by 2.
(y(x)/cosx)'=y2/cosx (3)

integrating equation 3.
y(x)/cosx=y2∫secx=y2ln|secx+tanx|+c (4)

isolating y in equation 4.
y(x)=cosxy2ln|secx+tanx|+c(cosx) (4)
Here is where I get stuck i don't know how to isolate for y.

## The Attempt at a Solution

.[/B]

#### Attachments

• IMG_0103.jpg
23.9 KB · Views: 615
• IMG_7594.JPG
26.4 KB · Views: 658
Last edited:
garr6120 said:

## Homework Statement

I am having trouble proving if the equation i have found for number 1 is correct. I have posted my solution to get back to the main problem in the first photo below.

For number 2 I am having trouble isolating for 1 y(x). Did i do the integration and setup properly?

## Homework Equations

As you can see y'+ycotx should equal 2cosx but when proven it equals 2cosx+sin-3xcosx

## The Attempt at a Solution

View attachment 211963 View attachment 211964 [/B]
Your posted photo is unreadable. Just type out the problem, so we can all see clearly what you want to do.

garr6120 said:

## Homework Statement

I am having trouble proving if the equation i have found for number 1 is correct. I have posted my solution to get back to the main problem in the first photo below.

For number 2 I am having trouble isolating for 1 y(x). Did i do the integration and setup properly?

## Homework Equations

Question 1:

given y' + ycotanx+2cosx (1) and y(π/2)=0 find the IVP.

find the integrating factor from equation 1.
μ(x)=e∫cotanxdx=sinx (2)

multiply equation 2 by 1.
y'sinx+ysinxcotanx=2cosxsinx
(sinxy)'=2cosxsinx (3)

integrate equation 3.
sinxy=∫2cosxsinx=sin2x+c (4)

Isolate y in 4.
y=sinx+c(csc2x)

******************************************************************
This is incorrect. From ##y \sin x = c + \sin^2 x## we get ##y = c/\sin(x) + \sin(x) = \sin(x) + c \csc(x)## (not ##\csc^2(x)##).

*******************************************************************

plug in the initial value to find c which is found to be -1.
y=sinx-csc2x (5)

proving that this is a solution to the differential:
take the derivative of equation 5.
y'=cosx+2csc2xcotanx

plug y and y' into differential equation 1.
y'+ycotanx=cosx+2csc2xcotanx+cosx-cotanxcsc2x
2cosx+3csc2xcotanx

I cannot get an answer for 2cosx. I did everything right.

**************************************************
No, you did not do everything right.

**************************************************

For Question 2:

y'+ytanx=y2 (1) for y(0)=1/2

find an integration factor.
μ(x)=e∫tanxdx=1/cosx {2}

multiply equation 1 by 2.
(y(x)/cosx)'=y2/cosx (3)

integrating equation 3.
y(x)/cosx=y2∫secx=y2ln|secx+tanx|+c (4)

***************************************************************************************************
This is incorrect. From ##(y/\cos x)' = y^2 / \cos x## you get ##y / \cos x = \int y(x)^2 / \cos x \; dx##. You cannot pull the ##y^2## outside the integral sign.

Integrating factors do not work for nonlinear differential equations, and yours is definitely nonlinear because it contains ##y^2##.

***************************************************************************************************isolating y in equation 4.
y(x)=cosxy2ln|secx+tanx|+c(cosx) (4)
Here is where I get stuck i don't know how to isolate for y.

.[/B]

## 1. What is an IVP?

An IVP, or initial value problem, is a type of mathematical problem that involves finding a solution to a differential equation based on a set of initial conditions. These initial conditions specify the value of the function at a particular point.

## 2. How do you solve an IVP?

The general process for solving an IVP involves using techniques from differential equations, such as separation of variables or the method of undetermined coefficients, to find a general solution. Then, the initial conditions are used to determine the specific solution that satisfies the given problem.

## 3. What is a First Linear Order IVP?

A First Linear Order IVP is a specific type of IVP where the differential equation involved is a first order linear differential equation. This means that the equation can be written in the form of y' + p(x)y = q(x), where p(x) and q(x) are functions of x.

## 4. How do you know if two First Linear Order IVPs are correct?

To determine if two First Linear Order IVPs are correct, you can solve each problem separately and compare the solutions. If the solutions are the same, then the IVPs are correct. You can also check that the initial conditions are satisfied by plugging in the values into the solution.

## 5. What are some real-world applications of solving IVPs?

IVPs have many applications in fields such as physics, engineering, and economics. For example, in physics, IVPs can be used to model the motion of objects under the influence of forces. In engineering, IVPs can be used to analyze and design systems such as circuits and control systems. In economics, IVPs can be used to model population growth and other dynamic systems.

• Calculus and Beyond Homework Help
Replies
8
Views
830
• Calculus and Beyond Homework Help
Replies
3
Views
649
• Calculus and Beyond Homework Help
Replies
21
Views
990
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
368
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
221
• Calculus and Beyond Homework Help
Replies
5
Views
391
• Calculus and Beyond Homework Help
Replies
10
Views
1K