For the 1st one, I’m not sure if I should define the curve as cos(Θ) sin(Θ) or x² + y². Disappointingly, neither does my tutor.
I should be able to solve it without too much bother if I can get a function to work from.

NFI where to start on this DE, any suggestions …? Or preferably just work it out so I can copy it please.

Spiny Norman said:

For the 1st one, I’m not sure if I should define the curve as cos(Θ) sin(Θ) or x² + y². Disappointingly, neither does my tutor.
I should be able to solve it without too much bother if I can get a function to work from.

NFI where to start on this DE, any suggestions …? Or preferably just work it out so I can copy it please.

I’d recommend parametrizing using Θ for the first one.

For the second one, this is a second order linear DE, kind of analogous to a quadratic equation …

dv said:

I’d recommend parametrizing using Θ for the first one.

For the second one, this is a second order linear DE, kind of analogous to a quadratic equation …

So that’s a sin/cos thingy for the 1st one then?
The NFI what to do with the 2nd one is unchanged.

Okay got the 1st one done, I think.
Came up with 24.

Okay gimme a minute

Spiny Norman said:

dv said:

I’d recommend parametrizing using Θ for the first one.

For the second one, this is a second order linear DE, kind of analogous to a quadratic equation …

So that’s a sin/cos thingy for the 1st one then?
The NFI what to do with the 2nd one is unchanged.

Okay, I’m not sure what methods you’ve been taught but this is one way to do it using D nomenclature, D representing differentiation.

We expect that a non-homogeneous second order linear differential equation will have a solution in two parts: one of them is the solution that would apply if the RHS was a zero (ie the homogeneous equation) y ₀, and a second part to allow for the RHS (y ₁). The second part is sometimes called the “particular solution”. The first part is sometimes called the “general solution to the homogeneous equation”.

Writing it in that D form would be

D²y + 2Dy – 3y = 3 e ^-t

(D² + 2D – 3 )y = 3 e ^-t

Factorise

(D + 3) (D – 1) y = 3 e ^-t

Setting aside the RHS for a moment, the roots of the quadratic on the left hand side are -3 and 1. This would give y ₀ = C ₀ e ^t + C ₁ e ^-3t

This leads to the solution C ₀ e ^t + C ₁ e ^-3t + y ₁

So we now need to find “the particular solution. The RHS has one of the exponents that we see in the general solution which is nice, so we’ll start there.

We assume that the particular solution will take the form C ₂ e ^-t

Subbing this into the original equation we get:

D²(C ₂ e ^-t) + 2D(C ₂ e ^-t) – 3(C ₂ e ^-t) = 3 e ^-t

bring that constant to the front

C ₂ ( D²( e ^-t) + 2D( e ^-t) – 3( e ^-t) )= 3 e ^-t

Using differentiation

C ₂ ( ( e ^-t ) + 2(- e ^-t ) – 3( e ^-t) )= 3 e ^-t

C ₂ = -3 /4

So we combine the “general” and “characteristic” parts

y = C ₀ e ^t + C ₁ e ^-3t + -3 /4 e ^-t

We now sub in what else we know to find the remaining variables.

We know y(0) = 0

So C ₀ + C ₁ = 3/4

and we know Dy(0) = 2 , and we can work out Dy = C ₀ e ^t + -3C ₁ e ^-3t + 3 /4 e ^-t

So C ₀ + -3C ₁ + 3 /4 = 2

C ₀ + -3C ₁ = 5/4

You can solve these simultaneous equations
4C ₁ = -1/2
C ₁ = -1/8
C ₀ = 7/8

Put those in:
y = 7/8 e ^t + -1/8 e ^-3t + -3 /4 e ^-t

Okay thanks for that. Unfortunately I still have no real idea what’s going on.

Spiny Norman said:

Okay thanks for that. Unfortunately I still have no real idea what’s going on.

Okay like I said I’m not sure what methods they’ve taught you but I can break this down into a few steps.

I think what I’ll have to do is just focus on all the other stuff in the course to make sure that I can do it, and maybe some of the basic DE’s, and ignore the rest. It might be enough to get me over the line overall.

Spiny Norman said:

I think what I’ll have to do is just focus on all the other stuff in the course to make sure that I can do it, and maybe some of the basic DE’s, and ignore the rest. It might be enough to get me over the line overall.

Okay well let me know if you want me to break this down.

https://www.youtube.com/watch?v=ly4S0oi3Yz8

Great video on partial differential equations. There is also an earlier video on ODEs.