linear differential equations

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May 13th 2008, 07:17 AM
samdmansam

linear differential equations

show that the differential eqn x^2 * dy/dx -y = 2x^3*e^(-1/x)
is linear, and solve it using the integrating factor method.

I have no idea how to even get started on this problem, if someone could show me in a few steps it would really help me out. I have alot more of these to do.
May 13th 2008, 08:05 AM
ThePerfectHacker

Quote:

Originally Posted by samdmansam

show that the differential eqn x^2 * dy/dx -y = 2x^3*e^(-1/x)
is linear, and solve it using the integrating factor method.

I have no idea how to even get started on this problem, if someone could show me in a few steps it would really help me out. I have alot more of these to do.

$y' - \frac{1}{x^2}y = 2xe^{-1/x}$ , let $p(x) = e^{\int -\frac{1}{x^2} dx}$ .

The solution is given by $y = \frac{1}{p(x)} \left\{ \int 2xp(x)e^{-1/x} dx + C \right\}$ .
May 13th 2008, 09:01 AM
PaulRS

Ok, suppose you have a differential equation of the form: $f'(x)+a(x)\cdot{f(x)}=b(x)$

Let $f(x)\cdot{e^{\int_c^xa(t)dt}}=q(x)$ (1) be an auxiliary function then: $f'(x)\cdot{e^{\int_c^xa(t)dt}}+a(x)\cdot{f(x)}\cdo t{e^{\int_c^xa(t)dt}}=q'(x)$

But note that: $\left(f'(x)+a(x)\cdot{f(x)}\right)\cdot{e^{\int_c^ xa(t)dt}}=b(x)\cdot{e^{\int_c^xa(t)dt}}=q'(x)$

Thus, by the FTC: $\int_c^xb(z)\cdot{e^{\int_c^za(t)dt}}dz=q(x)-q(c)=q(x)-f(c)$ check in (1) that $q(c)=f(c)$

Thus we have by (1) that: $e^{-\int_c^xa(t)dt}\cdot{\int_c^xb(z)\cdot{e^{\int_c^z a(t)dt}}dz}+f(c)\cdot{e^{-\int_c^xa(t)dt}}=f(x)$

This is the general solution to the first order equation

So suppose you are given $f(0)$ for example, then you should let $c=0$ above