Register to reply 
Please check if the differential equation is correct 
Share this thread: 
#1
Feb512, 05:35 AM

P: 1

I am practising for my test. The question is to solve a differential equation
dy/dx + y/x + 1 = 5x y(0) = 1. The answer that i have come up with is (xy+y)= 5x^3/3+5x^2/2+c by substituting the values x=0 and y=1 in to the general equation I get y(x+1)=5x^3/3 + 5x^2/2 +1 as the particular solution. Can you tell me how will the particular solution look like and why this particular solution exists? 


#2
Feb512, 06:53 AM

P: 4,573

For your equation your initial condition only effects a constant in the entire equation. Based on this what is the effect of adding or subtracting a constant in a general equation? (Hint: how does it 'shift' the function?) 


#3
Feb512, 04:47 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,553

The differential equation can be written [tex]\frac{dy}{dx}= 5x \frac{y}{x+1}[/tex] The function on the right side is differentiable for all y and all x except 1 so by the "fundamental existence and uniqueness theorem" for initial value problems, a unique solution to this problem exist for all x larger than 1. 


Register to reply 
Related Discussions  
Check Differential Equation Solution  Calculus & Beyond Homework  3  
Please check my work (differential equation)  Calculus & Beyond Homework  3  
Is this 2nd Differential Equation Answer Correct?  Calculus & Beyond Homework  8  
Correct way to solve this differential equation  Calculus & Beyond Homework  0  
Does this differential equation setup look correct?  Introductory Physics Homework  1 