 Hello. Namaste. Myself, Ms. Basargaon. Learning outcome. At the end of this session, students will be able to solve homogeneous linear differential equations. Now, solve x cube d cube by dx cube minus 2x dy by dx plus 4y equal to 0. Pause the video for a while and find the solution of this equation. I hope you have completed. Here, given equation is x cube d cube by dx cube minus 2x dy by dx plus 4y equal to 0, which is a clearly a homogeneous linear equation. Now, to solve the homogeneous equation, we have to put x equal to e raised to z that is z is equal to log x, so that x dy by dx is equal to dy, where d stands for dy dz and x cube d cube by dx cube is equal to d into d minus 1 into d minus 2 of y. Now, putting these values in the given equation, we get d into d minus 1 into d minus 2 minus 2d plus 4 of y equal to 0. That is by simplifying this expression, we get d cube minus 3d square plus 2d and minus 2d plus 4 of y equal to 0. That is d cube minus 3d square plus 4 of y equal to 0, which is a linear differential equation with constant coefficients and its auxiliary equation is f of d equal to 0 that is d cube minus 3d square plus 4e is equal to 0. Now, factors of this cubic polynomial are d plus 1 into d minus 2 whole square e is equal to 0. That is we get the roots as a d equal to minus 1, 2 and 2. Here, one of the roots is real and distinct. Remaining two roots are real and repeated. Therefore, the complementary function is given by Cf is equal to C1 e raised to minus z plus for repeated root C2 plus C3z into e raised to 2z. Now, here the right hand side of the given equation is 0. Therefore, the required solution is only Cf that is required solution is y equal to C1 e raised to minus z plus into bracket C2 plus C3z into e raised to 2z, which is the solution of the equation number 1. To get the solution of the original equation, we can put z is equal to log x that is x equal to e raised to z. We get the required solution of the given equation. Now, we will solve more different types of examples on homogeneous equations. You take example number 1. Solve x d square y by dx square minus 2 y by x e is equal to x plus 1 by x square. Now, if you look at this equation, here the equation is not homogeneous. It is not satisfying the definition of homogeneous equation. To get a homogeneous equation, multiply given differential equation both side by x. We get x into x that is x square d square y by dx square minus 2 into y by x into x. x will get cancelled that is you get minus 2 y which is equal to x into x that is x square plus 1 by x square into x that is plus 1 by x. Now, clearly the given equation is now reduced to homogeneous linear equation. To solve the homogeneous linear equation, again we can put x is equal to e raised to z that is z is equal to log x. So, that x d y by dx is equal to d of y and x square d square y by dx square is equal to d into d minus 1 of y where d stands for here d by dz. Now, substituting all these values in the given equation, we get d into d minus 1 that is the value of x square d square y by dx square minus 2 of y equal to e raised to 2 z plus 1 by e raised to z. By simplifying, we get d square minus d minus 2 of y is equal to e raised to 2 z plus e raised to minus z and let us call this as equation number 1. Now, which is a linear differential equation with constant quip sense and its auxiliary equation is given by f of d equal to 0 means d square minus d minus 2 is equal to 0. Now, factors of this quadratic polynomial are d plus 1 into d minus 2 equal to 0 that is, we get d equal to minus 1 and 2. Here, both the roots are real and distinct. Therefore, the complementary function can be written as C f is equal to C 1 e raised to minus z plus C 2 e raised to 2 z. Now, we will calculate P i. P i is equal to 1 by f of d. Here, f of d is d square minus d minus 2 of e raised to 2 z plus e raised to minus z. Now, we can separate into two parts which is equal to 1 by d square minus d minus 2 of e raised to 2 z plus 1 by d square minus d minus 2 of e raised to minus z. Now, here we can replace d by a means d by 2 here and if replaced by d by 2 here what we get 0. Therefore, we can multiply by z and we can differentiate the polynomial that is 1 by 2 d minus 1 of e raised to 2 z plus again here also we can multiply by z and derived the polynomial is 1 by 2 d minus 1 into e raised to minus z. Since, we know 1 by f of d of e raised to a x is equal to x into 1 by f dash of d of e raised to a x if f of a equal to 0 here f of a is 0. Therefore, we apply this rule. Now, which is equal to z into now we can replace here d by 2 that is 2 into 2 minus 1 of e raised to 2 z plus z into here we can replace d by minus 1 that is 2 into minus 1 minus 1 of e raised to minus z. Now, which is equal to z by 3 e raised to 2 z minus z by 3 e raised to minus z. Hence, the complete solution of the reduced equation that is equation number 1 is y equal to C f plus P i that is here y equal to C f is C 1 e raised to minus z plus C 2 e raised to 2 z plus this P i here we have taken z by 3 common z by 3 into bracket e raised to 2 z minus e raised to minus z. Now, to get the solution of original equation we can put z is equal to log x or e raised to z is equal to x we get y equal to C 1 e raised to minus z we will get x raised to minus 1 that is C 1 x raised to minus 1 plus C 2 x square plus this z as a log x that is log x by 3 into bracket x square minus x raised to minus 1 which is the solution of the given non homogeneous equation. Now, we will see one more example solve x cube d cube y by dx cube plus 2 x square d square y by dx square e is equal to x plus sin of log x. Now, clearly this equation is now homogeneous differential equation and reduce this homogeneous to linear we can put x equal to e raised to z that is z is equal to log x. So, that we get x square d square y by dx square is equal to d into d minus 1 of y and x cube d cube y by dx cube is equal to d into d minus 1 into d minus 2 of y substituting these values in the equation given equation becomes d into d minus 1 into d minus 2 plus this one 2 d into d minus 1 of y is equal to e raised to z plus sin z. Now, multiplying these two terms you get d cube minus 3 d square plus 2 d and multiplying this we get 2 d square minus 2 d of y is equal to e raised to z plus sin z that is d cube minus 3 d square plus 2 d square that is minus d square and plus 2 d minus 2 d you will get cancel of y is equal to e raised to z plus sin z. Now, which is a linear equation with constant coefficients now its auxiliary equation is given by d cube minus d square equal to 0 that is taking d square common d square into d minus 1 equal to 0 that is d equal to 0 0 and 1 here 0 0 is repeated. Therefore, corresponding to this c f we can write into brackets c 1 plus c 2 z and corresponding to the root 1 we can write c 3 e raised to z. Now, P i, P i is equal to 1 by f of d means d cube minus d square of e raised to z plus sin z which is equal to 1 by d cube minus d square of e raised to z plus 1 by d cube minus d square of sin z. Now, here we have separated into two parts let us call this first part as a P i 1 and this second part as a P i 2. Now, we will calculate P i separately first we will consider P i 1 that is equal to 1 by d cube minus d square of e raised to z. Now, here again we can apply the rule of 1 by f of d of e raised to a x here a is 1 if you replace a by 1 then would you get 1 minus 1 0 which is a case of failure for that we can multiply by z and we take the derivative of d cube minus d square that is 3 d square minus 2 d of e raised to z which is equal to z into now you can replace d by 1 that is 1 by 3 into 1 square minus 2 into 1 of e raised to z which is equal to z into e raised to z. Now, we will calculate P i 2 P i 2 is 1 by d cube minus d square of sin z which is equal to we can write d cube as a 1 by d into d square minus d square of sin z that we can write which is equal to 1 by d into minus 1 minus of minus 1 sin z since we know 1 by phi of d square of sin a x is equal to 1 by phi of minus a square of sin a x here a is 1 which is equal to d into minus 1 that is minus d minus into minus 1 that is plus 1 of sin z. Now, here to get in the denominator d square to apply the rule we can multiply this by 1 plus d and we can divide by 1 plus d that is 1 plus d divide by this minus d plus 1 is 1 minus d into 1 plus d of sin z is equal to 1 plus d divided by 1 minus d into 1 plus d is 1 minus d square of sin z which is equal to 1 plus d again replacing d square by minus a square you get 1 minus of minus 1 which is equal to 1 by 2 now we are operating 1 plus d on sin z 1 into sin z is sin z plus d of sin z here d stands for d by d z which is equal to 1 by 2 sin z and d of sin z is that is derivative of sin z is cos z hence the complete solution of equation number 1 is we have y equal to c f plus pi 1 and pi 2 now putting z is equal to log x or e raise to z equal to x we get to add a solution of the given equation.