Asked by Claire
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.
Answers
Answered by
MathMate
Please confirm or correct typo:
Using 3(x-3)(x^2-6x+23)^<b>(1/2)</b> as the answer to differentiating f(x)=(x^2-6x+23)^(3/2)
So
f'(x)=3(x-3)(x^2-6x+23)^<b>(1/2)</b>
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.
Using 3(x-3)(x^2-6x+23)^<b>(1/2)</b> as the answer to differentiating f(x)=(x^2-6x+23)^(3/2)
So
f'(x)=3(x-3)(x^2-6x+23)^<b>(1/2)</b>
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.
Answered by
Mgraph
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
(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
Answered by
MathMate
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.
In fact, 3) belongs to a different sentence. However, the process of solution does not change.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.