0
$\begingroup$

I was just doing a question from my textbook. I am practising for my test. The question is to solve a differential equation

$\displaystyle\frac{dy}{dx}+ \frac{y}{x} + 1 = 5x$, $y(0) = 1.$

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?

  • 0
    Your question is not clear. When you ask what the solution looks like, do you mean, what does the graph of $y(x+1)=5x^3/3+5x^2/2+1$ look like? And when you ask why this solution exists, I have no idea what you are asking.2012-02-05
  • 0
    hey this is the question in my textbook Find the particular solutions to the following initial value problems. In each case sketch a graph of your particular solution and give a brief reason why this particular solution exists.2012-02-05
  • 0
    Well, I'm sorry, I really don't know what the "why" question means, but for the graph, surely if you are learning about differential equations, you have already learned something about sketching graphs? My advice (assuming your answer is correct) is to divide both sides by $x+1$, do the polynomial long division on the right side to get (some polynomial) plus (some-number-over-$x+1$), and then use whatever you have learned about graphing functions.2012-02-05
  • 0
    The differential equation isn't defined at $x = 0$. In fact, none of its solutions can satisfy the initial condition you presented. Also, your solution isn't correct. Try plugging it into the equation to see why. Are you sure you didn't make a mistake while typing the problem in the question?2012-02-05
  • 0
    the question is dy/dx + (y/x+1)=5x2012-02-05
  • 0
    Do you still want help on this problem?2012-02-05
  • 0
    When you write $y/x+1$, do you mean $(y/x)+1$? or do you mean $y/(x+1)$? Everyone has been assuming the first, but since (as people point out) there are inconsistencies with that, maybe you mean the second?2012-02-06
  • 0
    Say, Rohit...two days...three answers...no word from you...still there?2012-02-08

3 Answers 3