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I need a bit of help with integrating factors in differential equations in the following question. Bit confused how to rearrange to get the correct format needed: $$(1+x^2)\frac{dy}{dx}-\frac{4x^3y}{1-x^2} = 1$$

Use the integrating factor method to find the general solution and show that: $$y=\frac{k+3x-x^3}{3(1-x^4)}$$

Thanks in advance.

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The correct form for an integrating factor is: $$\frac{dy}{dx}+P(x)y=Q(x) \tag{1}$$ Where $\mu(x)=e^{\int P(x)~dx}$ is the integrating factor.


To obtain the correct form, divide both sides by $1+x^2$ to obtain: $$\frac{dy}{dx}-\frac{4x^3}{(1-x^2)(1+x^2)}y=\frac{1}{1+x^2}$$ Expanding gives: $$\frac{dy}{dx}-\frac{4x^3}{1-x^4}y=\frac{1}{1+x^2}$$ Putting it in the form of $(1)$: $$\frac{dy}{dx}+\frac{4x^3}{x^4-1}y=\frac{1}{1+x^2}$$ Hence, one should use the integrating factor: $$\mu(x)=e^{\int \frac{4x^3}{x^4-1}~dx}$$ Can you continue?

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  • $\begingroup$ I end up getting to (x^4 - 1)(dy/dx) + y = (x^2 - 1). I am not sure if this is right as i then have a problem integrating. This is after multiplying by the integrating factor. $\endgroup$ – MathsRookie Mar 22 '17 at 19:50
  • $\begingroup$ Almost. If we multiply by: $$\mu(x)=x^4-1$$ You should obtain: $$(x^4-1)\frac{dy}{dx}+4x^3 y=x^2-1$$ Hence, it follows that: $$\frac{d}{dx}((x^4-1)y)=x^2-1$$ Now, integrating both sides w.r.t $x$, you should obtain: $$(x^4-1)y=\int (x^2-1)~dx$$ Which I am sure you can do. $\endgroup$ – projectilemotion Mar 22 '17 at 19:54

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