Sorry to post this again, but I am still unable to understand it and need help. Please help.1) Using 3(x-3)(x^2-6x+23)^2 as the answer to differentiating f(x)=(x^2-6x+23)^3/2, which I have been able to do, I need to find the general solution of the differential equation dy/dx= (2/27)*(x-3)*((x^2-6x+23)/(y))^1/2, (y>0) in implicit form. (* means to multiply)2) I need the particular solution of the differential equation in(1) for which Y=2 when X=1. 3)I then need this particular solution in explicit form (please state this). I know this is asking a lot but I am really struggling and close to despair with this. Thank you so much.
Please confirm or correct typo:
Using 3(x-3)(x^2-6x+23)^(1/2) as the answer to differentiating f(x)=(x^2-6x+23)^(3/2)
So
f'(x)=3(x-3)(x^2-6x+23)^(1/2)
For
dy/dx= (2/27)*(x-3)*((x^2-6x+23)/(y))^1/2
Separate variables and integrate:
∫ y^(1/2) dy = ∫ (2/27)(x-3)(x^2-6x+23)^1/2 dx
After integration,
(2/3)y^(3/2) = (2/81)(x^2-6x+23)^(3/2)+C
which reduces to
y^(3/2)=1/27(x^2-6x+23)^(3/2)+C
Substitute initital conditions x=1.3, y=2 to solve for
C=0.25755360984321
Therefore
y=(1/27(x^2-6x+23)^(3/2)+0.25755360984321)^(2/3)
for which (1.3,2) is a particular solution.
Check the solution by substituting y into the original differential equation and make sure the equation is satisfied for all values of x whenever y>0.
1)f'(x)=(1/2)*3(x^2-6x+23)^2*(2x-6)=
(1/2)*2(x-3)*3(x^2-6x+23)^2=
(x-3)*3(x^2-6x+26)=3(x-3)(x^2-6x+23)
In the differential equation we separate
variables:
27*sqrt(y)dy=2*(x-3)*sqrt(x^2-6x+23)dx
Let z=x^2-6x+23 then dz=(x^2-6x+23)'dx
dz=(2x-6)dx=2*(x-3)dx
27sqrt(y)dy=sqrt(z)dz Integrating
27*y^(3/2)/(3/2)=z^(3/2)/(3/2) + C
18*y^(3/2)=(2/3)*(x^2-6x+23)^(3/2) + C
2)If y=2 when x=1
18*2^(3/2)=(2/3)*18^(3/2) + C
In left side:
(2/3)*18^(3/2)=(2/3)*9^(3/2)*2^(3/2)=
(2/3)*27*2^(3/2)=18*2^(3/2) => C=0
The particular solution:
18*y^(3/2)=(2/3)*(x^2-6x+23)^(3/2)
27*y^(3/2)=(x^2-6x+23)^(3/2)
9^(3/2)*y^(3/2)=(x^2-6x+23)^(3/2)
9*y=x^2-6x+23
y=(x^2-6x+23)/9
3) 9y-x^2+6x-23=0
Sorry, MGraph is right. I misread the data as Y=2 and X=1.3 instead of Y=2 and X=1.
In fact, 3) belongs to a different sentence. However, the process of solution does not change.
I understand that you're struggling with these questions. I'll guide you step by step through each of them.
1) To find the general solution of the given differential equation, dy/dx = (2/27)*(x-3)*((x^2-6x+23)/(y))^1/2 (y>0), in implicit form, we need to separate the variables and integrate.
First, let's start by multiplying both sides of the equation by y^(1/2) to get rid of the fractional exponent:
y^(1/2) * dy/dx = (2/27)*(x-3)*(x^2-6x+23)^1/2
Next, we can rewrite (x^2-6x+23)^1/2 as (x-3)(x^2-6x+23)^1/2 because we have that term in our original equation (3(x-3)(x^2-6x+23)^2). So, our equation becomes:
y^(1/2) * dy/dx = (2/27)*(x-3)(x-3)*(x^2-6x+23)^1/2
Now, we can integrate both sides with respect to x:
∫ y^(1/2) * dy = ∫ (2/27)*(x-3)*(x-3)*(x^2-6x+23)^1/2 dx
Integrating the left side is straightforward:
(2/3) * y^(3/2) = ∫ (2/27)*(x-3)*(x-3)*(x^2-6x+23)^1/2 dx
To solve the integral on the right side, we can let u = x^2 - 6x + 23:
(2/3) * y^(3/2) = ∫ (2/27)*(x-3)*(x-3) * u^(1/2) dx
Now, differentiate u with respect to x to find du:
du/dx = 2x - 6
Rearrange the equation to solve for dx:
dx = (1 / (2x - 6)) du
Substitute this value of dx back into the integral:
(2/3) * y^(3/2) = ∫ (2/27)*(x-3)*(x-3) * u^(1/2) * (1 / (2x - 6)) du
= (2/27) ∫ (x-3)*(x-3) * u^(1/2) * (1 / (2x - 6)) du
Integrate the right side with respect to u:
(2/3) * y^(3/2) = (2/27) ∫ (x-3)*(x-3) * (2x - 6)^(-1) * u^(1/2) du
Now, substitute u back with its original value, x^2 - 6x + 23:
(2/3) * y^(3/2) = (2/27) ∫ (x-3)*(x-3) * (2x - 6)^(-1) * (x^2 - 6x + 23)^(1/2) du
Solving this integral would be quite complex, involving trigonometric substitutions. But if you proceed with the integration, you will eventually reach the general solution of the differential equation in an implicit form.
2) To find the particular solution of the differential equation when Y = 2 and X = 1, we need to substitute these values into the general solution and solve for the constant of integration.
Using the general solution obtained in step 1, substitute Y = 2 and X = 1:
(2/3) * 2^(3/2) = (2/27) ∫ (1-3)*(1-3) * (2*1 - 6)^(-1) * (1^2 - 6*1 + 23)^(1/2) du
Simplify the equation:
(4/3) = (2/27) ∫ (-2)^(-1) * 18^(1/2) du
Integrate the right side:
(4/3) = (2/27) ∫ ((-2)^(-1) * 18^(1/2)) * du
= (2/27) * (-2) * 2 * u^(1/2)
= (-4/27) * u^(1/2) + C
Now, substitute back the original value of u:
(4/3) = (-4/27) * (x^2 - 6x + 23)^(1/2) + C
3) To determine the particular solution in explicit form, we need to isolate the (x^2 - 6x + 23)^(1/2) term and solve for y:
(4/3) + (4/27) * (x^2 - 6x + 23)^(1/2) = y^(3/2)
Let's square both sides of the equation to eliminate the fractional exponent:
(4/3)^2 + 2 * (4/3) * (4/27) * (x^2 - 6x + 23)^(1/2) + (4/27)^2 * (x^2 - 6x + 23) = y^3
Now, simplify the equation:
(16/9) + (8/27) * (x^2 - 6x + 23)^(1/2) + (16/729) * (x^2 - 6x + 23) = y^3
Let's denote (x^2 - 6x + 23) as u again for clarity:
(16/9) + (8/27) * u^(1/2) + (16/729) * u = y^3
Finally, take the cube root of both sides to solve for y:
y = ((16/9) + (8/27) * u^(1/2) + (16/729) * u)^(1/3)
This is the particular solution of the given differential equation in explicit form when Y = 2 and X = 1.
I hope this explanation helps you with your problem. If you have any further questions, please let me know.