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

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

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

## Answer This Question

## Related Questions

- socials - ms. sue i actual still need help on first question because i read ...
- Dear Jiskha - Dear Jiskha, I really need help with some things and I am able to ...
- re -same question again - sorry to post this question again but i am having ...
- MATH - sorry I posted this question before. I have been ill.I still need help. ...
- ALGEBRA again still need help ! - need help solving this word problem, ...
- MATH - how many milligrams are there in a kilogram I give up alejandra. I showed...
- MsSue- re - im sorry my post appeared twice it didnt show my question when i ...
- To gina - I have deleted your post. PLEASE do not put personal information (...
- English - I urgently need you to check this. Thank you, Writeacher. 1) Sorry for...
- Physics-repost, sorry, I still don't understand :( - A forearm with a mass of 1....

More Related Questions