00:01
Hello students, to solve the given differential equation using the method of frobenius, we assume a power series solution of the form that is y of x is equal to summation cn into x to the power n plus r where cn are constants and r is the root of the initial, initial equation and n is non -negative integer.
00:40
Now we need to find the initial equation, the first bit.
00:48
Let's find the initial equation by substituting the power series solution into the differential equation and equating coefficients of like power of x.
00:57
So y dash is equal to summation cn into n plus r into x to the power n plus r minus 1.
01:07
Y double dash is equal to summation cn n plus r into n plus n r minus 1 into x to the power n plus r minus 2.
01:17
Now substitute these derivatives on the power series solution to the differential equation we will get 2x square into summation cn into n plus r into n plus r minus 1 into x to the power n plus r minus 2 minus x into summation cn n plus r into x to the power n plus r minus 1 plus x square plus 1 into summation cn into x to the power n plus r which is equal to 0.
01:55
Simplifying we will get summation 2cn n plus r into n plus r minus 1 into x to the power n plus r minus summation cn n plus r into x to the power n plus r plus summation cn x square plus 1 into x to the power n plus r equal to 0.
02:24
Grouping terms with same power of x we have summation 2cn into n plus r into n plus r minus 1 minus cn 2cn n plus r n plus r minus 1 minus cn n plus r plus cn into x to the power n plus r plus summation cn into x to the power n plus r plus 2 equal to 0.
02:54
Equating the coefficients of x to the power n plus r to the 0 give us the initial equation that is 2cn into n plus r into n plus r minus 1 minus cn into n plus r plus cn which is equal to 0.
03:13
The coefficient of cn cannot be 0 for all values of n therefore the bracketed term must be 0.
03:20
So, simplifying we will get 2 into n plus r n plus r minus 1 minus n plus r plus 1 equal to 0.
03:31
Simplifying this quadratic equation we will get 2n square plus 4r minus 2 into n plus 2r square minus 2r plus 1 equal to 0.
03:43
This is the initial equation.
03:47
This is initial equation.
03:55
The roots of this equation will give us value of now for the second bit finding linearly independent solution to obtain two linearly independent solutions we need to consider two distinct values of r...