00:01
The two solutions are given as e power minus 2x, sine 3x.
00:09
Obviously, the solution should be of the form.
00:14
A .e.
00:15
Power minus 2x plus b, sine 3x plus c cos 3x.
00:21
Because when sine 3x is a solution, cost 3x will also be a solution, because the differential equation is a linear differential equation with constant coefficients.
00:31
Now, i can write this as ae power minus 2x plus some lamb, lambda into e power 3i x to sum mu into e power minus 3 i x where 3 i is square root of minus 1 and by the virtue of yoiler's identity e power i theta is cost theta plus i sine theta that means the powers of e coefficients of x in the powers of e are negative to 3i negative 3 i so what will be the differential equation it will be d minus of minus 2 into d minus 3 i into d minus 3 i into d plus i is equal to 0 that is d plus 2 into d square plus 9 is equal to 0 or d cube plus 2 d square plus 90 is 18 equal to 0 this is the auxiliary polynomial so that means the differential equation will be d cubed y by d x cube plus 2 into d square y by d x square is 9 into d .y by d x is 18 y is equal to 0.
01:52
This is the third of a differential equation.
01:58
Y double dash plus y is equal to 6xxx.
02:05
That means this is d square plus 1 of y equal to 6 c power x.
02:11
Let's find a complementary solution.
02:13
D square plus 1 is 0.
02:15
D is equal to negative i and positive y.
02:17
So the complementary solution y c of x is a cosine of x is b sine x because a power i x is cost x x x x x where a and we have some arbitrary constants and particular integral is 1 by d square plus 1 of 6 e power x now we know that 1 by f of d e power lambda x is 1 by f of lambda he power lambda x where f of lambda is not equal to so that means 6 outside 1 by d square plus 1 e power x so it means 6 e power x divided by 1 square plus 1 so 3 power x so the final solution is y of x is a cos x plus b sine x plus 3 power x that's it this is the general solution the next equation is y double dash plus 4y dash plus 4y is equal to 5x e power negative 2 x so the solution is i mean the auxiliary polynomial is d squared plus 4 d plus 4 is equal to 0 that means it is d plus 2 into d plus 2 is equal to 0 so the complementary solution will be a x plus b into e power negative to x this is the complementary solution now let us find a particular integral is basically given by yp of x is 1 by d plus 2 whole square 5 into x e power negative to x how you can pull it out 5 into 1 by d plus 2 whole square x into e power negative to x now we have a formula 1 by f of d of x into x into a function of x that is x into v is x into 1 by f of t of v minus f dash of d divided by f of d by f of d of b so 1 by f of d of x e power minus 2 x is x into 1 by f of d power minus 2 x minus minus f dash of d divided by f of d whole square of d power minus 2 x what is f of d in our case it is d plus 2 whole square so 1 by d plus 2 whole square d power negative 2 x minus what is the f dash of the 2 into d plus 2 d plus 2 divided by f f f of d whole square means d plus 2 whole power 4 e power negative to x now what is 1 by d plus 2 e power negative to x it will be x into e power negative to x now we have very important property that 1 by f of d e power a x g of x is e power a x is e power a 1 by f of d plus a g of x so using this 1 by d plus 2 whole square e power negative 2 x x is d power negative 2 x 1 by i should replace d with what d minus 2 plus 2 whole square of x that is e power minus 2 x into 1 by d square of x what is 1 by d square of x double integration of x single integration is x squared by 2 one more integration is x cubed by 6 so it is x cube e power negative 2x divided by 6 but outside there is a 5 so the particular solution is 5 x cube by 6 e power negative 2 x so final answer for the solution is a x plus b e power negative 2 x plus 5 x cubed divided by 6 e power negative 2 x this is the final solution where a and we are arbitrary constants.
07:30
The y -dash minus 5i is equal to 3e power x minus 2x plus 1.
07:42
The auxiliary polynomial is d minus 5 is 0...