# calculus

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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.

• calculus - ,

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.

• calculus - ,

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

• calculus typo - ,

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.