Find the differential equation of this function and indicate the order y = e^3x (acos3x +bsin3x)​

y"-6y'+18y=0

Second order

Step-by-step explanation:

Since there are 2 constants, the order of the differential equation will be 2. This means we will need to differentiate twice.

y = e^(3x) (acos3x +bsin3x)​

y'=3e^(3x) (acos3x+bsin3x)

+e^(3x) (-3asin3x+3bcos3x)

Simplifying a bit by reordering and regrouping:

y'=e^(3x) cos3x (3a+3b)+e^(3x) sin3x (3b-3a)

y"=

3e^(3x) cos3x (3a+3b)+-3e^(3x) sin(3x) (3a+3b)

+3e^(3x) sin3x (3b-3a)+3e^(3x) cos(3x) (3b-3a)

Simplifying a bit by reordering and regrouping:

y"=

e^(3x) cos3x (9a+9b+9b-9a)

+e^(3x) sin3x (-9a-9b+9b-9a)

Combining like terms:

y"=

e^(3x) cos3x (18b)

+e^(3x) sin3x (-18a)

Let's reorder y like we did y' and y".

y = e^(3x) (acos3x +bsin3x)

y=e^(3x) cos3x (a) + e^(3x) sin3x (b)

Objective is to find a way to combine or combine constant multiples of y, y', and y" so that a and b are not appearing.

Let's start with the highest order derivative and work down

y"=

e^(3x) cos3x (18b)

+e^(3x) sin3x (-18a)

We need to get rid of the 18b and 18a.

This is what we had for y':

y'=e^(3x) cos3x (3a+3b)+e^(3x) sin3x (3b-3a)

Multiplying this by -6 would get rid of the 18b and 18a in y" if we add them.

So we have y"-6y'=

e^(3x) cos3x (-18a)+e^(3x) sin3x (-18b)

Now multiplying

y=e^(3x) cos3x (a) + e^(3x) sin3x (b)

by 18 and then adding that result to the y"-6y' will eliminate the -18a and -18b

y"-6y'+18y=0

Also the characteristic equation is:

r^2-6r+18=0

This can be solved with completing square or quadratic formula.

I will do completing the square:

r^2-6r+18=0

Subtract 9 on both sides:

r^2-6r+9=-9

Factor left side:

(r-3)^2=-9

Take square root of both sides:

r-3=-3i or r-3=3i

Add 3 on both sides for each:

r=3-3i or r=3+3i

This confirms our solution.

Another way to think about the problem:

Any differential equation whose solution winds up in the form y=e^(px) (acos(qx)+bsin(qx)) will be second order and you can go to trying to figure out the quadratic to solve that leads to solution r=p +/- qi

Note: +/- means plus or minus

So we would be looking for a quadratic equation whose solution was r=3 ×/- 3i

Subtracting 3 on both sides gives:

r-3= +/- 3i

Squaring both sides gives:

(r-3)^2=-9

Applying the exponent on the binomial gives:

r^2-6r+9=-9

Adding 9 on both sides gives:

r^2-6r+18=0

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )