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[clever dust]

Note that f'(x) = g(x) and g'(x) = -f(x) imply f''(x) = -f(x). So, you can solve that first with initial conditions f(2) = 4, f'(2) = 4.

[warm agate]

F"(x)+f(x)=0

But how to find the value of f(16)= ?

@clever dust

[clever dust]

You need to solve y'' + y = 0 first.

[warm agate]

C1e^(-x)+c2

[clever dust]

No.

[static cradle]

Use the characteristic polynomial then depending on the roots of it you have different solutions here you’ll have imaginary roots so the solutions are trigonometric

[warm agate]

I don't know what are you talking about?

[static cradle]

Do you know what the characteristic polynomial is for a linear homogeneous DE with constant terms ?

for example to solve y’’ -y’-y=0

Here the characteristic polynomial is r^2-r-1=0

Which has roots (1+-sqrt(5))/2

So here the general solution is then C1e^((1+sqrt(5))/2 *x) +C2e^((1-sqrt(5))/2 *x) with C1 and C2 two constants to determine with initial conditions

Here , you have y’’+y=0

So the characteristic polynomial is r^2+1=0

Which has imaginary roots i and -i

[warm agate]

What should I do with these roots?

[static cradle]

Well then the solutions are of the form C1e^ix+C2e^(-ix) which can be written more simply with eulers formula: Acos(x)+Bsin(x) with A and B being real constants

To determine with initial conditions

[warm agate]

A cos i + b sin i?

A cos i -b sin i

[static cradle]

no there is no i it’s just Acos(x) + Bsin(x)

So y(x)=Acos(x) +Bsin(x)

Now you can determine A and B with initial conditions

[static cradle]

What’s the question ?

[clever dust]

You need to identify how many groups of those five are non-abelian.

[static cradle]

Ok ty

[rapid echo]

@static cradle has given 1 rep to @clever dust

[static cradle]

Im a bit confused is it the order of all the groups combined that is 12 ?

Or are each of them of order 12

And are p and q positive integers ? Or do they have special properties, they can’t be simultaneously prime else p^2q can’t be even