00:01
So this is the given question we are asked to find the general solution of this differential equation.
00:06
Okay.
00:07
So let us write the actual differential equation from this.
00:11
This can be written as d square y by d x square minus 2d dy dyxx minus 8y equal to 4 e power 2x minus 2 .2x minus 8y equal to 4 e power 2x minus 21 e power minus 3x now this left side one is called the operator okay we can write this as d square minus 2d minus 8 times y equal to 4 e power 2x minus 21 e power minus 3 x here yeah d is nothing but d by d x okay this is d and d square is nothing but d square by d x squared similarly now so this can be written as f of d into y equal to 4 e power 2x minus 21 e power minus 3x so this is our differential equation we have transformed the given differential equation into this form now for this form the solution y has two parts one is complementary function and other one is particular integral p.
01:41
So the solution of y is complementary function plus particular integral.
01:46
You can also call it as yc complementary and yp particular.
01:51
Now let us try to find the both.
01:55
Now let us try to find the first one that is complementary cf.
02:01
To find yc, that is yc, okay? to find yc, f of d.
02:11
D should be equal to 0 we should equate the operator operator function f of d to equal to 0 and we should find the roots so it becomes d square minus 2d minus 8 equal to 0 okay so this is the operator f of d we are equating it to 0 now if you try to find the roots for this we can get okay so if you find the roots of this equation we get d equal to 4 comma minus 2 since the roots are real and distinct we can directly write a yc equal to suppose consider these roots are m1 and m2 we can directly write yc equal to c1 e power m1 x plus c2 e power m2 x so this becomes c1 e power 4x plus c to e power minus 2x okay so these are the solution for the complementary function y c now let us try to find a particular integral okay so this is the second part now to find a particular integral we have a technique particular integral p i we are finding y p okay to find y p we should operate this on the right side function that is uh so from this equation, see, from this equation, if you operate f of d, if we take f of d to the right side, then it becomes y equal to yp, this is nothing but the particular form, 1 by f of d, 4e power 2x minus 21e power minus 3x...