Help with Differential Equations please

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SUMMARY

This discussion focuses on solving two differential equations: \(\frac{dy}{dx}=3y\) with the condition \(f(2)=-1\) and \(\frac{dy}{dx}=e^{y}x\) when \(x=-2\) and \(y=-\ln(3)\). The user attempts to solve the first equation using integration but encounters issues with the natural logarithm of a negative number. The second equation is also addressed, with the user acknowledging a common mistake regarding the integral of \(1/y\) not being simply \(\ln(y)\) but rather \(\ln(|y|)\).

PREREQUISITES
  • Understanding of basic differential equations
  • Knowledge of integration techniques, specifically with logarithmic functions
  • Familiarity with the concept of initial conditions in differential equations
  • Basic calculus concepts from AP Calculus AB
NEXT STEPS
  • Review the properties of logarithmic functions, especially \(\ln(|y|)\)
  • Study methods for solving first-order differential equations
  • Practice solving differential equations with initial conditions
  • Explore the applications of differential equations in real-world scenarios
USEFUL FOR

Students in AP Calculus AB, particularly those struggling with differential equations, as well as educators looking for common pitfalls in teaching integration and logarithmic functions.

Jay9313
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Homework Statement


[itex]\frac{dy}{dx}[/itex]=3y f(2)=-1
and
[itex]\frac{dy}{dx}[/itex]=e[itex]^{y}[/itex]x when x=-2 y=-ln(3)
I'm in Calc AB By the way, so please do not try to show me methods that are too advanced.

Homework Equations


There are no relevant equations?


The Attempt at a Solution


My attempt at the first solution is
[itex]\int[/itex][itex]\frac{dy}{3y}[/itex]=[itex]\int[/itex]dx
[itex]\frac{ln(y)}{3}[/itex]=x+c
y=e[itex]^{3x+3c}[/itex]
ln(-1)=6+3c
You can't have that Natural Log though.. So I 'm stuck.

My attempt at the second solution is..
∫[itex]\frac{dy}{e^{y}}[/itex]=∫x dx
-[itex]\frac{1}{e^{y}}[/itex]=[itex]\frac{x^{2}}{2}[/itex]+c
-ln(e[itex]^{-y}[/itex])=e[itex]^{}\frac{x^{2}+2c}{2}[/itex]
-ln(3)=e[itex]^{}\frac{4+2c}{2}[/itex]
 
Last edited:
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Jay9313 said:

Homework Statement


[itex]\frac{dy}{dx}[/itex]=3y f(2)=-1
and
[itex]\frac{dy}{dx}[/itex]=e[itex]^{y}[/itex]x when x=-2 y=-ln(3)
I'm in Calc AB By the way, so please do not try to show me methods that are too advanced.

Homework Equations


There are no relevant equations?

The Attempt at a Solution


My attempt at the first solution is
[itex]\int[/itex][itex]\frac{dy}{3y}[/itex]=[itex]\int[/itex]dx
[itex]\frac{ln(y)}{3}[/itex]=x+c
Your integration has a common mistake. The integral of 1/y is NOT ln(y), it is ln(|y|).

y=e[itex]^{3x+3c}[/itex]
ln(-1)=6+3c
You can't have that Natural Log though.. So I 'm stuck.
(I'm going to submit this and go back and show my solution for the second one since it takes me a little while to put this in.)
(2) is also easy to integrate.
 
Oh man, I'm so used to the absolute value not mattering,that it screwed me up =/ Thank you soooo much.
 

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