Ordinary Differential equation

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SUMMARY

The discussion focuses on solving the first-degree ordinary differential equation (ODE) given by (5+7y²)dy/dx + (4x+2)y = 0 with the initial condition y(1)=1. Participants explored various methods including integrating factors and separation of variables, ultimately arriving at the implicit solution 5ln(y) + (7/2)y² = -2x² - 2x + C. The constant C was determined to be 15/2 using the initial condition. The importance of correct syntax when entering solutions into computational tools was emphasized, particularly the distinction between "ln(y)" and "lny".

PREREQUISITES
  • Understanding of first-degree ordinary differential equations (ODEs)
  • Familiarity with integration techniques, including integration by parts
  • Knowledge of implicit differentiation and initial value problems
  • Proficiency in using mathematical notation in computational software
NEXT STEPS
  • Study the method of integrating factors for solving ODEs
  • Learn about implicit differentiation and its applications in ODEs
  • Explore the use of computational tools for solving differential equations
  • Review common syntax requirements for mathematical expressions in programming languages
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as software developers working with mathematical computation tools.

cheddacheeze
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Homework Statement


Note: Here your solution is implicitly defined, i.e. you can not rearrange the solution to get an explicit expression for y. Therefore you need to enter your solution as an equation. You should enter your solution in the form f(x,y) = constant where you have determined both f(x,y) and the constant (for example: sin(y)/x=ln(2) ). You must include an equals sign. Do not use decimal numbers.



Homework Equations


(5+7y^2)dy/dx +(4x+2)y = 0, y(1)=1


The Attempt at a Solution


had a try at using integrating factor but having the equation = 0 doesn't really help
then tried breaking them up into seperable equations
((5+y^2)/y)dy = (-4x-2)dx
integrated and got f(x,y)=5lny+y^2-2x^2-2x+9/2

 
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(5+y^2)/y = (5/y) + y
when you integrate (5/y) + y , you should get 5Iny + (1/2)y^2
and when you shift your intergrated x components to the left side, shouldn't it be positive?
thats what i think..haha
 
hmm yea tried that aswell
also tried integrating:
(5+y^2)dy = (-4xy-y)dx
got 5y +(y^3)/3 = -4x^2y-2xy+c
with y(1)=1
got c = 34/3
but still wrong >.>
 
i don't think you can do it that way. You can't simply multiply the y over and integrate wrt x and treat y as constant.

opps! and sorry i read ur qns wrongly.
(5+7y^2)dy/dx +(4x+2)y = 0

this is 3rd degree DE? think i haven learn this yet..haha..sorry!
=P
 
its a first degree DE since it has dy/dx in it, not d^3y/dx^3
at least to my knowledge
 
cheddacheeze said:

Homework Statement


Note: Here your solution is implicitly defined, i.e. you can not rearrange the solution to get an explicit expression for y. Therefore you need to enter your solution as an equation. You should enter your solution in the form f(x,y) = constant where you have determined both f(x,y) and the constant (for example: sin(y)/x=ln(2) ). You must include an equals sign. Do not use decimal numbers.



Homework Equations


(5+7y^2)dy/dx +(4x+2)y = 0, y(1)=1


The Attempt at a Solution


had a try at using integrating factor but having the equation = 0 doesn't really help
then tried breaking them up into seperable equations
((5+y^2)/y)dy = (-4x-2)dx
That's a good start, but you made a mistake on the left side.
You should have (5/y + 7y)dy = (-4x - 2)dx

cheddacheeze said:
integrated and got f(x,y)=5lny+y^2-2x^2-2x+9/2
This is not quite right. You should have something like 5 ln(y) + <other x and y terms> = <constant>.
 
blursotong said:
you should get 5Iny + (1/2)y^2

There is no "In" function. That's Ln, which is sort of an abbreviation of natural Logarithm.
 
oh wow i can't believe i mistook the question, i didnt see 7y^2 while trying to work it out maybe that will help
 
this is the answer i got:
5logy + 7/2y^2 + 2x^2 + 2x + 15/2
but it still did not work
 
  • #10
That's because you lost your equation. Remember what I said...
You should have something like 5 ln(y) + <other x and y terms> = <constant>
Starting from here
(5/y + 7y)dy = (-4x - 2)dx
Integrate both sides of this equation to get
5lny + (7/2)y^2 = -2x^2 -2x + C

Now use the initial condition y(1) = 1 to find C, and you're done, so stop at that point.

If you want to check your answer, differentiate implicitly and you should get back to the original differential equation.
 
  • #11
Mark44 said:
That's because you lost your equation. Remember what I said...

Starting from here
(5/y + 7y)dy = (-4x - 2)dx
Integrate both sides of this equation to get
5lny + (7/2)y^2 = -2x^2 -2x + C

Now use the initial condition y(1) = 1 to find C, and you're done, so stop at that point.

If you want to check your answer, differentiate implicitly and you should get back to the original differential equation.

yes using the initial conditions
5ln1 + 7/2(1)^2 + 2(1)^1 +2(1) = C
which is
5(0) + 7/2 + 2 + 2 = C
C = 15/2 doesn't it?

which is the same equation i got and it still doesn't work
 
  • #12
Yes, I got C = 15/2. Why do you think it doesn't work?
 
  • #13
Mark44 said:
Yes, I got C = 15/2. Why do you think it doesn't work?

have tried entering it on the computer and it still is not right

sorry it was right, instead of entering "ln(y)", i entered as "lny" damn syntax
 
Last edited:
  • #14
Keep that in mind. We take a lot of shortcuts in writing math expressions, that a computer program is not likely to understand, especially with trig functions and log functions. Instead of cosx or sinx, it's probably going to be looking for cos(x) and sin(x).
 
  • #15
yep thanks for you help :)
 

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