Solving a 2nd Order DE: 6 & 7 Mark Questions

[whyyie]

[PLAIN]http://img213.imageshack.us/img213/7402/30788730.jpg
This questions consists of a total of 13 marks
- 6 marks for the first order DE
- 7 marks for the second order DE

I attempted the solution as below. I could solve the second order DE using normal method but when I think again the word "hence" does not allow me to do that. I notice that by differentiating the first order DE given, we can get second order DE. Is there a method to solve second order DE by integrating factor or any alternative solution?

[PLAIN]http://img138.imageshack.us/img138/6199/20128507.jpg

 

[Apphysicist]

Having limited experience with hyperbolic sine and cosine, I can't give you a relationship off the top of my head. But perhaps there is a relationship between 2x+sinh(2x) and 4*cosh(x)^2 that you may be able to exploit.

 

[whyyie]

Yup. By differentiating 2x+sinh(2x) we will be able to get 4*cosh(x)^2.
Also, the second order DE above is actually the derivatives of the first order DE.

But I didn't manage to figure out how to use the relationship between the first and second order DE to solve the question.

 

[rock.freak667]

Yup. By differentiating 2x+sinh(2x) we will be able to get 4*cosh(x)^2.
Also, the second order DE above is actually the derivatives of the first order DE.

But I didn't manage to figure out how to use the relationship between the first and second order DE to solve the question.

If you directly integrate the second DE, you will notice that the left side is the first DE.

\int \left( \frac{d^2y}{dx^2}+5 \frac{dy}{dx} \right) dx = \int 4cosh^2x dx

\Rightarrow \frac{dy}{dx} +5y = \int 4cosh^2x dx + C

So now work out the right side and see what you get .

 

[whyyie]

If you directly integrate the second DE, you will notice that the left side is the first DE.

\int \left( \frac{d^2y}{dx^2}+5 \frac{dy}{dx} \right) dx = \int 4cosh^2x dx

\Rightarrow \frac{dy}{dx} +5y = \int 4cosh^2x dx + C

So now work out the right side and see what you get .

Integrate that will get 2x + sinh (2x) with c = 0. i don't think that we can compare.

 

[whyyie]

I have try another method of using integrating factor for the second order DE.
But still I think there is nothing related with the word "hence" from the first part.
I have double check the answer, the answer satisfy the equation.

Is the word "hence" means continue from the answer from the first part?

[PLAIN]http://img338.imageshack.us/img338/474/64111659.jpg

 

[rock.freak667]

Why are you comparing the answers for?

 

[Apphysicist]

Integrate that will get 2x + sinh (2x) with c = 0. i don't think that we can compare.

Take a real hard look at what you quoted and what you said you get when you integrate.

I'm pretty sure that the integrating factor method only works with linear, first-order DEs. (dy/dx + f(x)*y=g(x)...mu=e^int(f(x)dx))

 

[hunt_mat]

There are some thing that maybe important:
<br /> \sinh 2x=2\cosh x\sinh x\quad\cosh 2x=2\cosh^{2}x-1\quad\sinh 2x=\frac{e^{2x}-e^{-2x}}{2}<br />
Use integrating factors on the first order ODE.