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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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    Say, Rohit...two days...three answers...no word from you...still there?2012-02-08

3 Answers 3

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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 :

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