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Amber's questions at Yahoo! Answers regarding linear first order ODEs

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  • #1


Staff member
Feb 24, 2012
Here are the questions:

Find the general solution of the differential equation?

(a) (x^2+1)dy/dx + xy = x

(b) (x + xy^2)dx + e^(x^2)ydy = 0

I don't think (b) is separable therefore there is no general solution. Is that correct to assume that because of the (x + xy^2) in brackets? Also I am not sure how to go about doing part (a).

Thanks! :)
I have posted a link there to this thread so the OP can view my work.
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  • #2


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Feb 24, 2012
Hello Amber,

We are asked to find the general solution to the following ODEs:

a) \(\displaystyle \left(x^2+1 \right)\frac{dy}{dx}+xy=x\)

One way to proceed is to separate the variables to obtain:

\(\displaystyle \frac{1}{1-y}\,dy=\frac{x}{x^2+1}\,dx\)

We should note that in doing so, we are losing the trivial solution $y\equiv1$.

We may integrate as follows:

\(\displaystyle -\int\frac{1}{y-1}\,dy=\frac{1}{2}\int\frac{2x}{x^2+1}\,dx\)

\(\displaystyle -\ln|y-1|=\frac{1}{2}\ln\left(x^2+1 \right)+C\)

Applying logarithmic properties, we may write (where we express the constant of integration as the natural log of a positive constant with no loss of generality):

\(\displaystyle \ln\left|\frac{1}{y-1} \right|=\ln\left(C\sqrt{x^2+1} \right)\)

Equating log arguments, we obtain:

\(\displaystyle \frac{1}{y-1}=C\sqrt{x^2+1}\)

Note: because of the absolute value of the argument on the left, the parameter $C$ may now be any non-zero value.

Inverting both sides, we obtain:

\(\displaystyle y-1=\frac{1}{C\sqrt{x^2+1}}\)

\(\displaystyle y(x)=\frac{1}{C\sqrt{x^2+1}}+1\)

We may redefine the parameter $C$ such that it is the inverse of that given above, and we may allow it to be zero to incorporate the trivial solution we lost during the separation of variables, and so we may state:

\(\displaystyle y(x)=\frac{C}{\sqrt{x^2+1}}+1\)

Another way we could solve the given ODE is to divide through by $x^2+1>0$ to obtain the linear ODE in standard form:

\(\displaystyle \frac{dy}{dx}+\frac{x}{x^2+1}y=\frac{x}{x^2+1}\)

Computing the integrating factor, we find:

\(\displaystyle \mu(x)=e^{\int\frac{x}{x^2+1}\,dx}=\sqrt{x^2+1}\)

Multiplying through by this factor, the ODE becomes:

\(\displaystyle \sqrt{x^2+1}\frac{dy}{dx}+\frac{x}{\sqrt{x^2+1}}y=\frac{x}{\sqrt{x^2+1}}\)

Observing the left side is now the differentiation of a product, we may write:

\(\displaystyle \frac{d}{dx}\left(\sqrt{x^2+1}y \right)=\frac{x}{\sqrt{x^2+1}}\)

Now we may integrate:

\(\displaystyle \int\,d\left(\sqrt{x^2+1}y \right)=\frac{1}{2}\int\frac{2x}{\sqrt{x^2+1}}\,dx\)

Computing the anti-derivatives, we obtain:

\(\displaystyle \sqrt{x^2+1}y=\sqrt{x^2+1}+C\)

Dividing through by \(\displaystyle \sqrt{x^2+1}\) we find:

\(\displaystyle y(x)=\frac{C}{\sqrt{x^2+1}}+1\)

b) \(\displaystyle \left(x+xy^2 \right)\,dx+e^{x^2}y\,dy=0\)

We may separate variables to obtain:

\(\displaystyle \frac{y}{y^2+1}\,dy=-e^{-x^2}x\,dx\)

We may integrate as follows:

\(\displaystyle \int\frac{2y}{y^2+1}\,dy=\int e^{-x^2}(-2x)\,dx\)

Computing the anti-derivatives, we find the implicit solution:

\(\displaystyle \ln\left(y^2+1 \right)=e^{-x^2}+C\)