 At the end of this session, students will be able to solve Legendary Linear Differential Equation. Now, first we will see what is Legendary Linear Equation. An equation of the form A x plus b raise to n into d raise to n y by dx raise to n plus k 1 into A x plus b raise to n minus 1 into d raise to n minus 1 y by dx raise to n minus 1 and so on plus k suffix n minus 1 into bracket A x plus b dy by dx plus k n y equal to x where k 1, k 2 and so on k n are constants and x is function of x. Now, such equations can be reduced to linear equation with constant coefficients by the standard substitution A x plus b equal to e raise to z that is z is equal to log of A x plus b. Then dy by dx will become dy by dz into dz by dx. As you know, dz by dx is 1 by A x plus b into A that is equal to dy by dz into A upon A x plus b that is we can write A x plus b into dy by dx equal to A into d of y here d stands for dy dz. Similarly, A x plus b whole square d square over dx square equal to we get A square into d into d minus 1 of y and A x plus b raise to 3 into d cube y by dx cube is equal to A cube into d into d minus 1 into d minus 2 of y and so on. After making this substitution in equation number 1, we get a linear differential equation with constant coefficients. Now, question is reduce the following differential equation into linear differential equation with constant coefficients. Pause the video for a while and reduce this given equation to linear differential equation with constant coefficients. I hope you have completed. Now, here given equation is 5 x plus 2 whole square into d square by dx square plus into bracket 5 x plus 2 dy by dx plus 10 y equal to 6 x. This is a Legendre linear differential equation. Comparing with Legendre's equation here A is 5 and b equal to 2 put 5 x plus 2 equal to raise to z that is z is equal to log of 5 x plus 2. So, that 5 x plus 2 dy by dx equal to 5 into d of y and 5 x plus 2 whole square d square y by dx square equal to A square means 25 d into d minus 1 of y. Then the given equation becomes 25 d into d minus 1 plus 5 d plus 10 of y equal to 6 into substituting the value of x that is e raise to z minus 2 by 5 that is 25 d square minus 20 d plus 10 of y equal to 6 into e raise to z minus 2 whole divided by 5. Now, we will solve the complete example solve 2 x plus 3 whole square into d square over dx square plus 2 x plus 3 into dy by dx minus 2 y equal to 24 x square. This is a clearly Legendre's equation comparing with Legendre's equation here A is 2 and b is equal to 3 put 2 x plus 3 equal to e raise to z that is z is equal to log of 2 x plus 3. So, that 2 x plus 3 into dy by dx is equal to 2 into d of y and 2 x plus 3 whole square d square y by dx square equal to 4 into d into d minus 1 of y substituting these values then the given equation becomes 4 into d into d minus 1 plus 2 d minus 2 of y is equal to 24 substituting the value of x that is e raise to z minus 3 by 2 whole square that is simplifying this we get 4 d square minus 4 d plus 2 d minus 2 of y is equal to 24 by 2 square means that is 4 into e raise to z minus 3 whole square that is 4 d square minus 4 d plus 2 d that is minus 2 d minus 2 of y is equal to 6 into taking the square of this we get e raise to 2 z minus 6 e raise to z plus 9. Now, dividing throughout by 2 we get 2 d square minus d minus 1 of y is equal to 3 into bracket e raise to 2 z minus 6 e raise to z plus 9. Now, this is a linear differential equation with constant coefficient and it is auxiliary equation is 2 d square minus d minus 1 equal to 0 and it is roots are d equal to 1 plus r minus square root of 1 plus 8 by 4 that is 1 plus r minus 3 by 4 that is 4 by 4 and minus 2 by 4 that is d equal to 1 and d equal to minus half. Therefore, the complementary function is given by c f is equal to c 1 e raise to z plus e raise to minus 1 by 2 z. Now, we will calculate the particular integral particular integral is what 1 by f of d here f of d 2 d square minus d minus 1 of 3 into e raise to 2 z minus 6 e raise to z plus 9 which is equal to 3 into we can separate into 3 parts 1 by 2 d square minus d minus 1 of e raise to 2 z minus 6 into 1 by d 1 by 2 d square minus d minus 1 of e raise to z plus 9 into 1 by 2 d square minus d minus 1 of e raise to 0 z. Now, here by using the formula of 1 by f of d of e raise to a x and if you put the value of d as a 2 we get 1 by 2 into 4 minus 2 minus 1 of e raise to 2 z minus 6 z into 1 by 4 d minus 1 of e raise to z plus 9 into 1 by 0 minus 0 minus 1 of e raise to z which is equal to 3 into bracket that is 8 minus 3 1 by 5 e raise to 2 z minus 6 z into e raise to z by 3 and by simplifying this we get minus 9 therefore, particular integral is equal to 3 by 5 into e raise to 2 z minus 6 z into e raise to z minus 27. Then the general solution is given by y equal to c f plus p i here c f is c 1 e raise to z plus c 2 e raise to minus z by 2 plus 3 by 5 into e raise to 2 z minus 6 z into e raise to z minus 27. Now, replacing z by log up to x plus 3 we get y equal to c 1 e raise to log up to x plus 3 plus c 2 into e raise to minus log up to x plus 3 divided by 2 plus 3 by 5 e raise to twice log up to x plus 3 minus 6 into log up to x plus 3 into e raise to log up to x plus 3 minus 27 by simplifying this we get y equal to c 1 into 2 x plus 3 plus c 2 into 2 x plus 3 raise to minus 1 by 2 plus 3 by 5 into 2 x plus 3 whole square minus 6 into 2 x plus 3 into log up 2 x plus 3 minus 27. Now, we will see one more example solve x plus 1 whole square d square y by d x square plus x plus 1 d y by d x plus y is equal to log up x plus 1 raise to the power 4 plus cos of log up 1 plus x here comparing we get a equal to 1 and b equal to 1 now we have to put x plus 1 equal to e raise to z that is z d is equal to log up x plus 1 so that x plus 1 d y by d x equal to d up y and x plus 1 whole square d square y by d x square equal to d into d minus 1 of y substituting these values in the given equation we get d into d minus 1 plus d plus 1 of y is equal to z raise to 4 plus cos z that is by simplifying this we get that is d square minus d plus d plus 1 of y equal to z raise to 4 plus cos z that is d square plus 1 of y equal to z raise to 4 plus cos z. Now, it is auxiliary equation is given by d square plus 1 equal to 0 that is d equal to plus or minus i here both roots are imaginary therefore, you get c up is equal to c 1 cos z plus c 2 sin z now p i is equal to 1 by d square plus 1 of z raise to 4 plus cos z now here we have to separate into two parts that is 1 by d square plus 1 of z raise to 4 plus 1 by d square plus 1 of cos z we denote first one as a p i 1 second one as a p i 2 now first we calculate p i 1 that is equal to 1 by d square plus 1 of z raise to 4 because it is a polynomial therefore, you can shift d square plus 1 to the numerator that is d raise d square plus 1 raise to minus 1 of z raise to 4 which is equal to exponent in this by using binomial expansion 1 minus d square plus d raise to 4 minus d raise to 6 and so on of z raise to 4 which is equal to 1 minus d square plus d raise to 4 of z raise to 4 because of z raise to 4 here we have considered up to d raise to 4 which is equal to now operating this bracket on z raise to 4 1 into z raise to 4 is z raise to 4 minus d square of z raise to 4 plus d raise to 4 of z raise to 4 now that is equal to z raise to 4 d square of z raise to 4 is minus 12 z square and d raise to 4 of z raise to 4 is derivative of 4 of z dash to 4 is 24. Now, we will calculate P i 2 which is equal to 1 by d square plus 1 of cos z which is equal to z into 1 by 2 d of cos z because here if you replace d square by minus 1 it is a case of failure. Therefore, we can multiply by z and we can differentiate d square plus 1 with respect to d that is 1 by 2 d of cos z which is equal to z by 2 1 by d of cos z means integration of cos z t z which is equal to z by 2 into integration of cos z is sin z. Therefore, the complete solution is given by y equal to C a plus P i that is C 1 cos z plus C 2 sin z plus P i that is z dash to 4 minus 12 z square plus 24 and plus z by 2 sin z. Now, replacing z by log of x plus 1 we get y equal to C 1 cos of log of x plus 1 plus C 2 sin of log of x plus 1 plus log of x plus 1 is to 4 minus 12 log of x plus 1 whole square plus 24 plus log of x plus 1 by 2 into sin of log of x plus 1.