Please help. I have an implicit answer to a differential equation of 18y*sqrt(y)=(2(x^2-6x+23)^3/2)/3 plus arbitrary constant. What is the particular solution to this for which y=2 when x=1. And then express this particular solution answer in an explicit form. Please help and many thanks.

Choose the value of the arbitary constsnt C so that y = 2 when x = 1.

That is all there is to it.

36sqrt2 = 2*[2*18^(3/2)]/3 + C
= 2*(4/3)27*sqrt2) + C
= 72 sqrt2 + C

C = -36 sqrt2

You can write the explicit solution for y(x) by taking the 2/3 root of y^(3/2)/18, with the C term inserted

Hello drwls. Please can you explain how you get 72sqrt2+c. Surely, if it is 50.911=50.911+c, if one subtracts either side c=0? Am I missing something?Thank you for your help.

To find the particular solution to the given differential equation, we can use the method of separation of variables. However, since the equation is presented implicitly, we first need to simplify it to an explicit form.

Let's start by rearranging the equation to isolate the square root term with respect to y:

18y * sqrt(y) = (2(x^2-6x+23)^(3/2))/3 + C (where C is the arbitrary constant)

Squaring both sides of the equation, we get:

(18y * sqrt(y))^2 = ((2(x^2-6x+23)^(3/2))/3 + C)^2

Simplifying further, we have:

324y^3 = (2(x^2-6x+23)^(3/2))^2/9 + 2C((x^2-6x+23)^(3/2))/3 + C^2

Now, let's substitute the given values y = 2 and x = 1 into the equation to solve for C:

324(2)^3 = [(2(1^2-6(1)+23)^(3/2))^2]/9 + 2C((1^2-6(1)+23)^(3/2))/3 + C^2

Simplifying the equation, we get:

2592 = [2(18)^(3/2))^2]/9 + 2C(18)^(3/2))/3 + C^2

2592 = [2(18)^(3/2)^2]/9 + 2C(18)^(3/2))/3 + C^2

2592 = (2(324))/9 + 2C(18))/3 + C^2

2592 = 72 + 12C + C^2

Rearranging the equation, we have:

C^2 + 12C + 72 = 2592

C^2 + 12C - 2520 = 0

Now, we can solve this quadratic equation for C using factoring, quadratic formula, or any other preferred method. Once we obtain the value(s) of C, we can plug it back into the original equation, simplify, and express the particular solution in an explicit form.