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Help rearranging a linear first order differential equation
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- Jan 26, 2012
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Re: Help rearranging a linear first order differential
\[\frac{dy}{dx}-\frac{8}{x}y=3x^2-2x+2.\]
We now note that the equation is now in the form of a linear equation $\dfrac{dy}{dx}+P(x)y=Q(x)$. To proceed from here, you need to compute the integrating factor
\[\mu(x)=\exp\left(\int P(x)\,dx\right)=\ldots\quad(\text{I leave this part to you})\]
where $\exp(x)=e^x$. Then if you multiply both sides of the linear ODE by $\mu(x)$, you get
\[\frac{d}{dx}[\mu(x) y]=\mu(x)(3x^2-2x+2)\implies y=\frac{1}{\mu(x)}\int \mu(x)(3x^2-2x+2)\,dx.\]
Can you fill in the work I left out? I hope this helps!
The first thing you need to do is get every term involving y on one side of the equation. So subtracting $\dfrac{8}{x}y$ from both sides gives youHi I am trying to solve dy/dx = 3x^2-2x+2+(8/x *y)
Can anyone show me how to rearrange to standard form as I am mightly confused
\[\frac{dy}{dx}-\frac{8}{x}y=3x^2-2x+2.\]
We now note that the equation is now in the form of a linear equation $\dfrac{dy}{dx}+P(x)y=Q(x)$. To proceed from here, you need to compute the integrating factor
\[\mu(x)=\exp\left(\int P(x)\,dx\right)=\ldots\quad(\text{I leave this part to you})\]
where $\exp(x)=e^x$. Then if you multiply both sides of the linear ODE by $\mu(x)$, you get
\[\frac{d}{dx}[\mu(x) y]=\mu(x)(3x^2-2x+2)\implies y=\frac{1}{\mu(x)}\int \mu(x)(3x^2-2x+2)\,dx.\]
Can you fill in the work I left out? I hope this helps!
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