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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?

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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. –  Gerry Myerson Feb 5 '12 at 11:35
    
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. –  Rohit Pradeep Feb 5 '12 at 11:42
    
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. –  Gerry Myerson Feb 5 '12 at 11:51
    
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? –  Ayman Hourieh Feb 5 '12 at 12:21
    
the question is dy/dx + (y/x+1)=5x –  Rohit Pradeep Feb 5 '12 at 12:32
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3 Answers

The differential equation is undefined at $x=0$, so we may as well assume $x\ne 0$ and multiply by $x$ no harm done. Then we get $$xy' + y + x = 5x^2\ ,$$ which we can rewrite as $$(x y)'= 5x^2-x\ .$$ A function $x\mapsto y(x)$ is a solution of this equation iff $$x\ y(x)={5\over3} x^3 -{1\over 2}x^2 + C\qquad(x\ne 0)$$ for some real constant $C$, or $$y(x)={5\over3} x^2 -{1\over 2}x + {C\over x}\qquad (x\ne0)\ .$$ It is impossible to fulfill the condition $y(0)=1$ even in its weaker form $\lim_{x\to 0} y(x)=1$ with any choice of $C$.

PS: I don't know how you found "$(xy+y)= 5x^3/3+5x^2/2+c$"; in any case it is erroneous.

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Assuming that your equation is really $${dy\over dx}+{y\over x+1}=5x$$ (and not ${dy\over dx}+{y\over x}+1=5x$), your solution is correct, so, if I understand you correctly, you want to know what the graph of $$y={{5x^3\over3}+{5x^2\over2}+1\over x+1}$$ looks like. As noted in my comment, you can do polynomial long division to rewrite this in the form $$y=A(x)+{B\over x+1}$$ for some quadratic polynomial $A(x)$ and some number $B$. When you get to that stage, if you need help sketching the graph, you can always ask.

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Rewrite equation into form :

$y'+\frac{1}{x}y=5x-1$

This is First Order Linear Differential Equation so general solution is given by :

$$y = \frac {\int u(x) \cdot (5x-1) \,dx+C}{u(x)} ~~\text {where}~~ u(x)= e^{\int \frac{1}{x} \,dx}$$

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