Integrating factor of $xy'+ 4y= x^2-x+1$

Question

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:

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

Answer 1

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

Answer 2

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.

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