 Hello, Namaste Myself, M.S. Basargaon, Assistant Professor, Department of Humanities and Sciences, Walsh and NISTOR Technology, Solapur. Now, learning outcome. At the end of this session, students will be able to solve homogeneous linear equation. Now, we will see what is Cauchy's homogeneous linear equation. An equation of the four x raise to n d raise to n y by dx raise to n plus k 1 x raise to n minus 1 into d raise to n minus y by dx raise to n minus 1 plus so on plus k suffix n minus 1 into x dy by dx plus k n y equal to x, where k 1, k 2, k 3 and so on k n are constants and x is function of x is called Cauchy homogeneous linear equation. Now, such equation can be reduced to linear differential equation with constant coefficients by using the substitution x equal to e raise to z, that is z d is equal to log x. Then we can calculate the value of dy by dx, that is dy by dx is equal to dy by dz into dz by dx. Here, dz by dx is 1 by x, which is equal to dy by dz into 1 by x. Multiplying throat by x, you get that is x dy by dx is equal to dy, where here d stands for dy dz. And differentiating both side dy by dx with respect to x, we get similarly x square d square y by dx square is equal to d into d minus 1 of y. Similarly, we get x cube d cube y by dx cube is equal to d into d minus 1 into d minus 2 of y and so on. After making the substitution in equation number 1, that is in Cauchy homogeneous equation, we get a linear differential equation with constant coefficient, which can be solved by the method of linear differential equation with constant coefficients. Now, solve the following linear differential equation with constant coefficient d square y by dx square minus phi u dy by dx plus 6 y equal to e raise to 3x. Pause the video for a while and find the solution of this equation. Now, I hope you have solved here given equation d square y by dx square minus phi u dy by dx plus 6 y equal to e raise to 3x. And in symbolic form, it can be written as d square minus phi u d plus 6 of y equal to e raise to 3x. And here, we have to calculate first CF, that is with the help of auxiliary equation. The auxiliary equation is here, d square minus phi u d plus 6 equal to 0. Now, factors of this polynomial are d minus 2 into d minus 3, that is d minus 2 into d minus 3 equal to 0, that is d equal to 2 and d equal to 3. Now, we can write the complementary function as CF is equal to c 1 e raise to 2x plus c 2 e raise to 3x. Now, we can calculate pi, equal to 1 by d square minus phi u d plus 6 of e raise to 3x, which is equal to x into 1 by 2d minus phi of e raise to 3x. Since, we know 1 by f of d of e raise to ax is equal to x into 1 by f dash of d e raise to ax if f of a equal to 0. Here, a is 3, if you put d equal to 3, here the denominator will become 0. Now, which is equal to x into 1 by, now we can replace d by 3 here, that is 2 into 3 minus phi e raise to 3x, which is equal to x into e raise to 3x. Therefore, the complete solution of the given linear differential equation is y equal to c f plus p r, that is y equal to here c f is c 1 e raise to 2x plus c 2 e raise to 3x and pi is plus x e raise to 3x. Now, we will solve the homogeneous differential equations, solve x square d square y by dx square minus 2x dy by dx minus 4y is equal to x raise to 4. Now, this is a homogeneous linear equation and to express this homogenous to linear, the standard substitutions are x equal to e raise to z, that is z is equal to log x. So, that x dy by dx is equal to d of y and x square d square y by dx square equal to d into d minus 1 of y, where here d stands for d by dz. Now, substituting all these values in the given equation, we get d into d minus 1 minus 2d minus 4 of y equal to e raise to z raise to the power 4, that is by simplifying this we get d square minus 3d minus 4 of y equal to e raise to 4z, let us call this as a equation number 1. Now, which is a linear differential equation with constant coefficients and its auxiliary equation is d square minus 3d minus 4 equal to 0, that is the factors of this polynomial are d plus 1 into d minus 4 equal to 0, that is d equal to minus 1 and 4. Here both the roots are real and different, therefore you can write cf as a equal to c1 e raise to minus z plus e raise to 4z. Now, we will calculate the second part, that is pi, pi equal to 1 by d square minus 3d minus 4 of e raise to 4z, again if you replace here d equal to 4 here, in place of d square and d we get 0, therefore here we can multiply by z, which is equal to z into 1 by derivative of d square is 2d and derivative of minus 3d is minus 3 and derivative of minus 4 is 0 of e raise to 4z, since 1 by f of d of e raise to ax is equal to x into 1 by f of d of e raise to ax, if f of e equal to 0, which is equal to no, we can replace here d by 4, that is z into 1 by 2 into 4 minus 3 of e raise to 4z, which is equal to z by 8 minus 3, that is 5 of e raise to 4z. Hence, the solution of equation number 1 is y equal to cf plus pi, that is equal to c1 e raise to minus z plus c2 e raise to 4z plus pi is z by 5 e raise to 4z. Putting z is equal to log x, that is e raise to z equal to x, we get y equal to c1 x raise to minus 1 plus c2 x raise to 4 plus log x by 5 x raise to 4, which is the solution of the given equation. Now, we will solve one more example, solve x square d square y by dx square plus 2x dy by dx minus 12y is equal to xq log x. Now, we can make use of the solution x equal to e raise to z, that is z is equal to log x, so that x dy by dx equal to d of y, x square d square y by dx square equal to d into d minus 1 of y, where d stands for d by dz. Now, substituting these values in the given equation, we get d into d minus 1 plus 2d minus 12 of y is equal to e raise to 3z into z. By simplifying this, we get d square plus d minus 12 of y equal to e raise to 3z into z, let us call this as a equation number 1 and which is the linear differential equation with constant coefficients. Now, its auxiliary equation is d square plus d minus 12 is equal to 0, that is factors of this polynomial are d plus 4 into d minus 3 equal to 0, that is the d equal to minus 4 and 3. Therefore, complementary function for these values, we get c of e is equal to c1 e raise to minus 4z plus c2 e raise to 3z. Therefore, now we can calculate p i, p i is equal to 1 by f of d, that is here f of d is d square plus d minus 12 of e raise to 3z into z, which is equal to, we can write e raise to 3z and we can replace d by d plus 3, that is d plus 3 whole square plus d plus 3 minus 12 of z, since we know 1 by f of d of e raise to ax into v is equal to e raise to ax into 1 by f of d plus a of v, which is equal to e raise to 3z 1 by d square plus 6d plus 9 plus d plus 3 minus 12 of z, that is equal to e raise to 3z into 1 by d square plus 7 d plus z. Now, by simplifying it, p i equal to e raise to 3z 1 by 7d, now shifting this to the numerator, we can write 1 plus d by 7 raise to minus 1 of z, which is equal to e raise to 3z by 7d. Now, expanding this by using binomial expansion, we get 1 minus d by 7 plus d by 7 whole square minus so on of z, since we know the binomial expansion 1 plus x raise to minus 1 is 1 minus x plus x square minus x cube and so on, which is equal to e raise to 3z by 7d. Now, operating this bracket on z, 1 into z, that is z minus 1 by 7 into d of d, which derivative of z is 1, that is 1 by 7, which is equal to e raise to 3z by 7 into bracket, what we get z minus 1 by 7 d z, which is equal to e raise to 3z by 7. Now, integration of this z minus 1 by 7 is z square by 2 minus 1 by 7 z, thus the solution of equation number 1 is y equal to c 1 e raise to minus 4z plus c 2 e raise to 3z plus p i, that is e raise to 3z by 7 into bracket z square by 2 minus 1 by 7 z. Now, putting z is equal to log x and e raise to z equal to x, we get y equal to c 1 x raise to minus 4 plus c 2 x cube plus x cube by 7 into bracket log x whole square by 2 minus 1 by 7 x, which is the solution of the d 1 equation.