- #1
coomast
- 279
- 1
Hello,
I have a problem using the program xmaxima. It involves the substitution of a new dependent and independent variable in an ordinary differential equation. Let me clearify this with an example of which we know the solution beforehand. So consider the following equation:
[tex]\frac{d^2y}{dx^2}+x \cdot y=0[/tex]
Substituting [itex]y=\sqrt{x}\cdot g(x)[/itex] gives:
[tex]x^2 \cdot \frac{d^2g}{dx^2}+x \cdot \frac{dg}{dx}+\left(x^3-\frac{1}{4}\right) \cdot g=0[/tex]
Substituting in this equation [itex]x^3=t^2[/itex], we get:
[tex]t^2 \cdot \frac{d^2g}{dt^2}+t \cdot \frac{dg}{dt}+\left(\left(\frac{2t}{3}\right)^2 -\left(\frac{1}{3}\right)^2\right) \cdot g=0[/tex]
Which is a Bessel differential equation with solution:
[tex]g(t)=A\cdot J_{1/3}\left(\frac{2t}{3}\right)+ B\cdot Y_{1/3}\left(\frac{2t}{3}\right)[/tex]
Transforming into the previous variables:
[tex]g(x)=A\cdot J_{1/3}\left(\frac{2}{3}x^{3/2}\right)+ B\cdot Y_{1/3}\left(\frac{2}{3}x^{3/2}\right)[/tex]
and thus the solution to the original differential equation:
[tex]y(x)=\sqrt{x}\cdot \left[A\cdot J_{1/3}\left(\frac{2}{3}x^{3/2}\right)+ B\cdot Y_{1/3}\left(\frac{2}{3}x^{3/2}\right)\right][/tex]
Now the question is how does one do that in xmaxima?
best regards,
coomast
I have a problem using the program xmaxima. It involves the substitution of a new dependent and independent variable in an ordinary differential equation. Let me clearify this with an example of which we know the solution beforehand. So consider the following equation:
[tex]\frac{d^2y}{dx^2}+x \cdot y=0[/tex]
Substituting [itex]y=\sqrt{x}\cdot g(x)[/itex] gives:
[tex]x^2 \cdot \frac{d^2g}{dx^2}+x \cdot \frac{dg}{dx}+\left(x^3-\frac{1}{4}\right) \cdot g=0[/tex]
Substituting in this equation [itex]x^3=t^2[/itex], we get:
[tex]t^2 \cdot \frac{d^2g}{dt^2}+t \cdot \frac{dg}{dt}+\left(\left(\frac{2t}{3}\right)^2 -\left(\frac{1}{3}\right)^2\right) \cdot g=0[/tex]
Which is a Bessel differential equation with solution:
[tex]g(t)=A\cdot J_{1/3}\left(\frac{2t}{3}\right)+ B\cdot Y_{1/3}\left(\frac{2t}{3}\right)[/tex]
Transforming into the previous variables:
[tex]g(x)=A\cdot J_{1/3}\left(\frac{2}{3}x^{3/2}\right)+ B\cdot Y_{1/3}\left(\frac{2}{3}x^{3/2}\right)[/tex]
and thus the solution to the original differential equation:
[tex]y(x)=\sqrt{x}\cdot \left[A\cdot J_{1/3}\left(\frac{2}{3}x^{3/2}\right)+ B\cdot Y_{1/3}\left(\frac{2}{3}x^{3/2}\right)\right][/tex]
Now the question is how does one do that in xmaxima?
best regards,
coomast