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Differential equation solved by Laplace transform

Posted March 21st, 2008 by Structure

Let begin with an example.
We want to solve initial value problem for the differential equation

$y"(x)+6y'(x)+25y(x)=f(x)u(x)$

with y(+0)=a and y'(+0)=b
Let

$Y(s)=\mathcal{L}(y(x))(s)=\int_{0}^{+\infty}y(x)e^{-sx}dx$

We have

$F(s)=\mathcal{L}(f(x)u(x))(s)$

$\mathcal{L}(y'(x))(s)=sY(s)-a;\:\mathcal{L}(y"(x))(s)=s^2Y(s)-as-b$

We have then

$s^2Y(s)-as-b+6(sY(s)-a)+25Y(s)=F(s)$

$(s^2+6s+25)Y(s)=as+6a+b+F(s);\:Y(s)=\frac{as+6a+b}{s^2+6s+25}+\frac{1}{s^2+6s+25}F(s)$

Now we can write

$Y(s)=\frac{a(s+3)}{(s+3)^2+16}+\frac{3a+b}{(s+3)^2+16}+\frac{1}{(s+3)^2+16}F(s)$

Derivative of implicite function.

Posted March 20th, 2008 by Isoscel

Let consider the equation $F(x,y)=x^2+y^2-1=0$
We know that the set of solution of this equation is the set of points in plane at distance 1 from the origin, or a circle of radius 1 with center in (0,0).
We want to give a description of this set depending on a single variable instead of two.
This is possible only local ,not for the whole set of solutions.
But the collection of local solutions can give us a complete information about the circle.
Let $(a,b)\in R^2$ with $a^2+b^2-1=0$ and $b\neq 0$
For $(x,y)$ in a small neighborhood of $(a,b)$ we have $y\neq0$