X^3fracd^3ydx^3 + 2x^2fracd^2ydx^2 - xfracdydx + y = 12x^2;...

Transcript

00:01 Hi, in this question we are given with the differential equation x cube times d cube y by dx cube plus 2x square times d square y by dx square minus x times dy by dx plus y is equals to 12 times x square and we need to check the general solution of this given differential equation for the given result.

00:27 For that if we start with the given differential equation to be replaced for the value of y to be equals to x raise to power lambda so we get x cube times d cube by dx cube times x raise to power lambda plus 2x square times d square by dx square times x raise to power lambda minus x times d dx of x raise to power lambda plus x raise to power lambda to be equals to 0 considering the characteristic equation only.

01:02 Solving this further we get this as replacing this third derivative second and first derivative we get x cube to be multiplied by lambda times lambda minus 2 lambda minus 1 times x raise to power lambda minus 3 plus 2x square times lambda times lambda minus 1 x raise to power lambda minus 2 minus x times lambda times x raise to power lambda minus 1 plus x raise to power lambda to be equals to 0.

01:42 Solving this further we get lambda cube minus lambda square minus lambda plus 1 and whole multiplied by x raise to power lambda to be equals to 0.

01:52 As solving this further for these as a factor we get lambda minus 1 square times lambda minus 1 to be equals to 0 that will give lambda to be equals to 1 1 and minus 1.

02:04 So for lambda to be equals to minus 1 we have y 1 x to be equals to c 1 by x for lambda to be equals to 1 as it is has the multiplicity of 2 so we get y 2 x to be equals to c 2 times x and y 3 x will be equals to c 2 by x.

02:26 So we get y 3 x to be equals to c 3 x times log of x.

02:30 So we get y of x to be equals to c 1 by x plus c 2 x plus c 3 x times log of x.

02:40 Now as we have the solution for y 1 y 2 and y 3 and now we can find the wronskian for the given differential equation where it will be equals to as 1 by x x x times log of x here its derivative as a first derivative that is minus 1 by x square 1 log of x plus 1 then here it will be having its second derivative that will be 2 by x cube 0 and 1 by x.

03:12 Now solving this further for its determinant we get the wronskian to be equals to 4 by x square...

Question

$x^3frac{d^3y}{dx^3} + 2x^2frac{d^2y}{dx^2} - xfrac{dy}{dx} + y = 12x^2$; ewline $y = c_1x^{-1} + c_2x + c_3xln{x} + 4x^2$

Question

$x^3frac{d^3y}{dx^3} + 2x^2frac{d^2y}{dx^2} - xfrac{dy}{dx} + y = 12x^2$; ewline $y = c_1x^{-1} + c_2x + c_3xln{x} + 4x^2$

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