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April 18, 2015

April 18, 2015

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

- calculus -
**MathMate**, Sunday, May 22, 2011 at 5:55pmPlease 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.

- calculus -
**Mgraph**, Sunday, May 22, 2011 at 6:16pm1)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 -
**MathMate**, Sunday, May 22, 2011 at 7:16pmSorry, 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.

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