Wondering if these two First Linear Order IVPs are correct

[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]

 

[Ray Vickson]

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.

 

[Ray Vickson]

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)

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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 $\csc^2(x)$).

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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.

The Attempt at a Solution

.[/B]