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# Differential equations watch

1. Er, a little stuck here.

Solve the following equation subject to the given boundary conditions:
z'' = -g (where g is a constant and z is a function of t.)

with i) z = 5 at t = 1 and z = 4 at t= 2

ii) z' = 0 at t = 2 and z = 0 at t = 3.

I don't know why but I actually don't know where to start here. This is analogous to acceleration due to gravity in physics and if we set z as position the solution should have some sort of z = 0.5gt2 somewhere but I honestly have no idea what to do otherwise...
2. Er, a little stuck here.

Solve the following equation subject to the given boundary conditions:
z'' = -g (where g is a constant and z is a function of t.)

with i) z = 5 at t = 1 and z = 4 at t= 2

ii) z' = 0 at t = 2 and z = 0 at t = 3.

I don't know why but I actually don't know where to start here. This is analogous to acceleration due to gravity in physics and if we set z as position the solution should have some sort of z = 0.5gt2 somewhere but I honestly have no idea what to do otherwise...
Start by rearranging like this

Now use the normal method of solving an ODE (auxillary equation, complimentary function etc).

Once this is done, you're ready to put in the conditions you've been given to find the values of the constants you will have.

Quote me with reply if you need mroe help!
3. (Original post by trm90)
...
It's just a straight integration (done twice); don't forget to add a constant of integration each time, and substitute initial conditions at the appropriate stage to determine the constants.
Start by rearranging like this

Now use the normal method of solving an ODE (auxillary equation, complimentary function etc).

Once this is done, you're ready to put in the conditions you've been given to find the values of the constants you will have.

Quote me with reply if you need mroe help!
(Original post by ghostwalker)
x
Thanks guys, I managed

I have one more question...

Find the solution of y'' + 2y' + 10y = 13cos2x subject to the initial conditions y(0)=0, y'(0)=2.

I'm not sure if I'm obtaining the general solution in the right way. If I am I should be able to find the final solution after imposing the boundary conditions, but any guidance would be nice!

First I found the auxiliary equation:

thus so the complementary factor is:

where a and b are arbitrary complex constants.

Then I had to find a p.i. for the equation...

I tried y = Acos2x + Bsin2x

Differentiated it, subbed everything back into the original equation to get:

(6A + 2B)cos2x + (6B - 4A)sin2x = 13cos2x

Comparing coefficients:

6A + 2B = 13
6B - 4A = 0

But then I get really ugly fractions for A and B. This is part of a multiple choice question and none of the available options have any sort of ugly fractions I found from those simultaneous equations.

Thanks again
5. (Original post by trm90)
6A + 2B = 13
6B - 4A = 0
Last one for the day.

The first one quoted is not correct; it' s not 2B.

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Updated: February 11, 2010
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