Differantial Equations

[Guest]

d[tex]y_{1 }[/tex]/dx=4y_{1 }+2y_{2 }+x

dy_{2 }/dx=-y_{1 }+y_{2 }

solve the system of equations.

can anyone help?

[HallsofIvy]

That's a system of differential equations. There are a number of different ways to solve such problems. What methods have you learned?

The first method I learned, not necessarily the simplest, is to differentiate the first equation again:
[tex]\frac{d^2y_1}{dx^2}= 4\frac{dy_1}{dx}+ 4\frac{dy_2}{dx}+ 1[/tex].

Replace [tex]\frac{dy_2}{dx}[/tex] using the second equation: [tex]\frac{d^2y_1}{dx^2}= 4\frac{dy_1}{dx}+ 4\left(-y_1+ y_2\right)+ 1[/tex]

[tex]\frac{d^2y_1}{dx^2}= 4\frac{dy_1}{dx}- 4y_1+ 4y_2+ 1[/tex]

From the first equation, again, [tex]4y_2= \frac{dy_1}{dx}- 4y_1[/tex] so
[tex]\frac{d^2y_`}{dx^2}= 4\frac{dy_1}{dx}- 4y_1+ \frac{dy_1}{dx}- 4y_1+ 1= 5\frac{dy_1}{dx}- 5y_1+ 1[/tex]/

The equation [tex]\frac{d^2y_1}{dx^2}- 5\frac{dy_1}{dx}+ 5y_1= 1[/tex].

The "associated homogeneous equation" is [tex]\frac{d^2ay_1}{dx^2}- 5\frac{dy_1}{dx}+ 5y_1= 0[/tex] and has characteristic equation [tex]r^2- 5r+ 5= 0[/tex] which has roots [tex]\frac{5}{2}\pm\frac{\sqrt{5}}{2}[/tex]. The general solution to the associated homogeneous equation is [tex]e{(5/2)x}\left(Ae^{(\sqrt{5}/2)x}+ Be^{-(\sqrt{5}/2)x}\right)[/tex]. To find a single solution to the entire equation, try a solution of the form y= A. Then [tex]\frac{d^2y}{dx^2}= \frac{dy}{dx}= 0[/tex] so the equation becomes [tex]-5A+ 1= 0[/tex] so [tex]A= \frac{1}{5}[/tex].

The general solution is [tex]y(x)= e^{(5/2)x}\left(Ae^{(\sqrt{5}x/2}+ Be^{-\sqrt{5}x/2}\right)+ \frac{1}{5}[/tex].

Now use [tex]4y_2= \frac{dy_1}{dx}- 4y_1[/tex] to solve for [tex]y_2[/tex].

[HallsofIvy]

A slightly more "sophisticated", more symbolic, is to replace the derivatives with "D" so that we can write the two equations as
[tex]Dy_1= 4y_1+ 4y_2+ x[/tex] and [tex]Dy_2= -y_1+ y_2[/tex] and then [tex](D- 4)y_1- 4y_2= x[/tex] and [tex]y_1+ (D- 1)y_2= 0[/tex].

Solve those as algebraic equations: "Multiply" the first equation by (D- 4) (actually differentiating) to get [tex](D- 1)(D- 4)y_1- 4(D- 1)x= (D- 4)x[/tex] or [tex](D^2- 5D+ 4)y_1- 4(D- 4)y_2= 1- 4x[/tex] and multiply the second equation by 4 to get [tex]4y_1+ 4(D- 4)y_2= 0[/tex] Adding the two equations eliminates [tex]y_2[/tex] and leaves [tex](D- 4)(D- 1)y_1+ Dy_1= (D^2- 5D+ 5)y_1= 0[/tex]. The "characteristic equation" is "[tex]D^2- 5D+ 5= 0[/tex]" which has roots, by the quadratic equation, [tex]D= \frac{5\pm\sqrt{5}}{2}[/tex]. So the general solution to the associated homogeneous equation is (as in the previous post) [tex]y= e^{(5/2)x}\left(Ae^{(\sqrt{5}/2)x}+ Be^{-(\sqrt{5}/2)x}\right)[/tex].