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How do I solve this linear differential equation?

$$xy'+ 4y= x^2-x+1.$$

I am supposed to use the integrating factor to solve it but I do not really understand how to implement it. I do know however, a linear first order O.D.E has the form:

$$\frac{\mathrm dy}{\mathrm dx} + P(x) y = Q(x).$$

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    $\begingroup$ By transposing terms and dividing/multiplying by some terms, being the expression you have into that of a linear ODE with the appropriate $P$ and $Q$. Remember $y'$ has to be isolated, so you have to divide by $x$ in this case. In fact, if you divide by $x$ now itself, you will get a linear ODE. Then you can solve it via the formula for linear ODE, calculating the integrating factor etc. $\endgroup$ – астон вілла олоф мэллбэрг Apr 25 '18 at 5:24
  • $\begingroup$ Note : If you have attempted the problem after reformulating as a linear ODE, you may post it as answer below. $\endgroup$ – астон вілла олоф мэллбэрг Apr 25 '18 at 5:45
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Multiply by $e^{\int P(x)dx}$ on both sides. Here $P(x)=\frac4x$.

So get $x^4$.

So $$\frac d{dx}(x^4y)=x^5-x^4+x^3$$.

Integrate both sides. Get $$ x^4y+C_1=\frac{x^6}6-\frac{x^5}5+\frac{x^4}4+C_2$$.

Now divide by the integration factor.

$$y=\frac{x^2}6-\frac x5+\frac14+\frac{C_3}{x^4}$$.

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  • $\begingroup$ I got an equation that is similar but on my LHS I have: I have x^4y' + 4(x^3)(y) $\endgroup$ – Jenny Apr 25 '18 at 6:11
  • $\begingroup$ Yes! Another way is to notice that if the RHS is a polynomial we have particular solutions with that are polynomials. Indeed $x y'+4y= (n+4)x^n$ if $y=x^n$. Btw, you forgot to divide $C_3$ by $x^4$. $\endgroup$ – Orest Bucicovschi Apr 25 '18 at 6:12
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    $\begingroup$ @Jenny that's correct... next note that, by the product rule, you have $\frac d{dx}(x^4y)$... $\endgroup$ – Chris Custer Apr 25 '18 at 6:20
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I followed your way and divided the whole equation by x

I now have the equation: y'+ $\frac 4x$y= x - 1 + $\frac 1x$ I then proceeded with I(x)= $e$$^{4logx}$

I then multiplied the whole equation by I(x) and got: $$x^4 y'+ 4x^3y = x^5 - x^4 + x^3$$

Since $e$$^{4logx}$ = $e^{logx^4}$ and e$^{log}$ cancels each other out

I'm kinda stuck again and dont really know how to continue

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