Posted by Claire on Sunday, May 22, 2011 at 1:33pm.
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.