Entered xln1 3x2 answer preview x ln1 3x2 result incorrect...

Entered $x*ln(|1-3*(x^2)|)$ Answer Preview $x ln(|1-3x^2|)$ Result incorrect The answer above is NOT correct. Consider the equation $(1-3x^2)y'' + 6xy' - 6y = 0$. Suppose that you somehow guessed that $y = x$ is a solution. Use the reduction of order method to find a second linearly independent solution, and solve the initial value problem with $y(0) = 3$, $y'(0) = 3$. $y=x ln(|1-3x^2|)$ help (formulas) Book: Section 2.1 of Notes on Diffy Qs

Transcript

00:01 Hello everyone in this problem we have given that is 4 x squared y double dash plus y is equal to 0 also we have given that is y 1 is equals to x raised 2 half into natural log of x and we have to check that it is a solution of this given equation so here the y1 is given so we will calculate the y dash that is equals to derivative of y with respect to x that is x raised to par half into natural log of x.

00:34 So after differentiating, we get 1x into x raised to par half plus 1 by 2 into 1 by x raised to par half into natural logo of x.

00:45 So after solving, we get by dash is equals to 1 by x raised to par half plus 1 by 2 x raise to par half into natural logo of x.

00:57 In the similar way, we will get divide double dash.

01:00 That is the derivative of single y dash so here we will get minus 1 by 2 into 1 by x raised 2 2 3 by 2 plus 1 by 2 x raised to part 3 by 2 plus 1 by 2 natural log of x into minus 1 by 2 x 3 by 2 so after solving we get y double dash is equals to minus 1 by 4 natural log of x into 1 by x raised to part 3 by 2.

01:40 Now in the next step we will just substitute the values of y dash in the given equation that is 4 x squared into y dash that is minus 1 by 4 natural log of x into 1 by x into 1 by x raised to part 3 by 2 plus y that is given x raise to par half into natural log of x so after solving we get here minus minus x x x x x to par half into natural log of x plus x x to par half natural log of x so here this one and this one both are same but having the opposite signs so it will be cancel out and comes out as zero which is equals to the right side of the given equation so we can say the y1 that is x raise to par 1 by 2 natural log of x is a solution of given equation.

02:47 Now the next part we have to find the second factor for the solution that is y2.

02:55 So by reducing method we will just consider that y2 x is equal to v of x.

03:01 Now we will get the y -dash x that is that is equals to v plus v dash into x.

03:10 In the similar way y double dash x comes out two times v dash plus v double dash x.

03:18 Now just substitute these two values in the given equation that is 4 x square into 2v dash plus v double dash x plus v of x is equal to 0.

03:33 So here after multiplying we get 4 times v double dash x cube plus 8 x square into v dash plus v of x is equals to 0.

03:46 Now just divide the equation by x we get 4x square v double dash plus 8 x v dash plus v is equal to 0.

03:58 Now we have to find the values of v so that we can get the value of v2 that is equals to v into x...

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