Question regards integral calculus

Hi its me again and I have return with more question that I currently got stuck on

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For the following differential equation

,

show that the substitution , yields the differential equation for

Hence find the solution to the original differential equation that satisfies the condition . Find the interval on which the solution to the initial value problem is defined.

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So I first set out to integrate first, however I have a problem that I do not know how to move to the otherside of the equation

e.g.

what I did is

I integrated by using integration by parts with u=tan(x) and

so the equation become

+ C

I'm certain that my integration is wrong, since y(0)=2 will result in

The same problem also applies to

since I do not know how to move, the and to the other side of the equation. So can any body tell me how to integrate this question?

Thanks

Junks

P.S. I finally posted this in latex :D

edit: I'll try to use First Order, O.D.E but since will first order O,D,E work there?

By first order O.D. E

I found my to be

is this correct?

Re: Question regards integral calculus

Well when you make the substitution then that means , so substituting into your DE you get

This is now first order linear and so can be solved using an integrating factor.

Re: Question regards integral calculus

Thank you, "Prove it" I got the idea of solving this problem now

Re: Question regards integral calculus

I have more question to ask in regards to this same question.

so I got

since then

I subbed this value of y into

the equation become

If I did not made any mistake in the process, how do you integrate this equation ?

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I think I need to integrate so i can solve for this

"Hence find the solution http://latex.codecogs.com/png.latex?y(x) to the original differential equation that satisfies the condition http://latex.codecogs.com/png.latex?y(0)=2. Find the interval on which the solution to the initial value problem is defined."

Re: Question regards integral calculus

Quote:

Originally Posted by

**junkwisch** If I did not made any mistake in the process, how do you integrate this equation

?

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

I think I need to integrate

so i can solve for this

"

Hence find the solution http://latex.codecogs.com/png.latex?y(x) to the original differential equation that satisfies the condition http://latex.codecogs.com/png.latex?y(0)=2. Find the interval on which the solution to the initial value problem is defined."

I haven't checked whether your working is correct, but use integration by parts to compute the above integral.

.

Recall .

Re: Question regards integral calculus

It should be du/dx = - tan x - sec x/2

not the du/dx = -u tanx - 2sec x given above.

you then take the onto the LHS and you now have a linear first order ODE.

integrating factor is sec x.

get down to u = -sinx/2 + C cos x

from y= sinx +1/u, u = 1/(y-sinx)

Im not sure where to go from here, can anyone help solving for y?

Re: Question regards integral calculus

I guess we didnt have exactly the same question

Im not sure if what i did is right, but try sub u into y then y into dy/dx

P.s. writing on my phone right now so I couldnt write a latex