Given y1(t) = t^2 and y2(t) = t^-1 satisfy the corresponding...

Given y1(t) = t^2 and y2(t) = t^-1 satisfy the corresponding homogeneous equation of t^2y'' - 2y = 2t^3 + 3, t > 0 Then the general solution to the non-homogeneous equation can be written as y(t) = c1y1(t) + c2y2(t) + Y(t). Use variation of parameters to find Y(t). Y(t) =

Transcript

00:01 Hi, let's start the solution.

00:01 In this question we have given t square y double dash minus 2y is 2 t q plus 3 t is positive.

00:11 This can be written as t squared d square minus 2 into y is 2 t q plus 3.

00:19 So this is a in the form of cauchy oiler equation.

00:26 So now put x equal to e power not x put t equal to e power z.

00:31 So, of.

00:32 Or z equal to log t and t squared d square equal to theta theta minus one by simplifying this we get theta square minus theta theta so here we get theta square minus theta minus two into y is two t cube is t power 3 z that is e power 3 z plus so now we know that for general solution is y of t is y c of t plus y p of t so for particular integral we use variation of parameter and for y c for y c of t complementary function write the auxiliary equation we get m square minus m minus 2 is 0 so here we get m minus 2 m plus 1 equal to 0 so that means here we have m is 2 and m is minus 1 so here roots are distinct so solution is y c of t is c1 e power m 1 t that is minus z plus c2 e power m 2 m 2 is so we get e z e power z square so now this is in the form of z z terms so now put the value of z here we get y c of t is c 1 is c 1 e power minus log t plus c2 e power z is log t squared and this can be written as e power 2 z.

02:29 Don't be confused.

02:31 So here we get 2 log t.

02:35 So by using log property here we get yc of t is c1e power log t inverse plus c2 e power log t square so that means here we get y c of t is c1 t inverse plus c2 t squared so here y1 is t inverse and y2 is t squared that is given so now we consider alternate in question we have given y 1 is t square and y2 is that is y and that is y2 y2 is the inverse now we find variation value of u by using variation of parameter so we get u is minus y2 r of t upon rwn skin of t d t so now we find the ron skin ron skin of t is y1 y2 that is t square and y2 is t power minus 1 differentiation of y 1 is 2 t of 1 by t is minus 1 upon t squared.

03:58 So determinant of 2 cross 2 matrix is a cross multiplication.

04:02 So here we get minus 1 upon t square into t square minus 2 t upon t...

Question

Given y1(t) = t^2 and y2(t) = t^-1 satisfy the corresponding homogeneous equation of t^2y'' - 2y = 2t^3 + 3, t > 0 Then the general solution to the non-homogeneous equation can be written as y(t) = c1y1(t) + c2y2(t) + Y(t). Use variation of parameters to find Y(t). Y(t) =

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