euler.html

Euler Equations

In the case that the functions , and of a differential equation

are not constants then it is much harder to solve it. In this section we consider the special case when , , and where and are constants. This equation is called Euler equation. First we solve the homogeneous equation , and then we use The Method of Variations to find a particular solution . Solutions of the homogeneous equation depend on the roots of the characteristic equation . If the roots and are real and different then the solution is . If the roots and are real and equal then the solution is . If the roots and are complex with and being the real and imaginary part then the solution is

. We ilustrate this on the following examples.

Example 1: Solve .

Solution: The roots of the characteristic equation are

> solve(r*(r-1)+2*r-12=0);0>solve(r*(r-1)+2*r-12=0);

real and different. Hence the solution is

Example 2: Solve .

Solution: The roots of the characteristic equation are

> solve(r*(r-1)-3*r+4=0);

equal. Hence the solution is .

Example 3: Solve .

Solution: The roots of the characteristic equation are

> solve(r*(r-1)+5*r+13=0);

complex with and . Hence the solution is

Example 4: Solve a nonhogenous Euler equation .

Solution: The roots of the characteristic equation are

> solve(r*(r-1)+2*r-12=0);

Hence two independent solutions of the homogeneous equation are and . We find a particular solution using the method of variation. Recall that for this method to work the differential equation has to be in the standard form . Thus new for Euler equation is equal to .

>

> y[1]:=x^(-4);y[2]:=x^3;

> solve({y[1]*dc[1]+y[2]*dc[2]=0,diff(y[1],x)*dc[1]+diff(y[2],x)*dc[2]=sqrt(x)/x^2},{dc[1],dc[2]});

> c[1]:=int(-1/7*x^(7/2),x);c[2]:=int(1/7/x^(7/2),x);

So the particular solution is

> c[1]*y[1]+c[2]*y[2];

Hence the general solution to the nonhomogeneous equation is .

Example 5: Solve

Solution: First we find by solving the equation .

> solve(r*(r-1)-2*r+2=0,r);

Therefore the solutions to the homogenous part are and . To find one particular solution of the original equation we use the method of variation as follows:

> y[1]:=x;y[2]:=x^2;

> dy[1]:=diff(y[1],x);dy[2]:=diff(y[2],x);

> solve({y[1]*dc[1]+y[2]*dc[2]=0,dy[1]*dc[1]+dy[2]*dc[2]=(3*x^2+2*ln(x))/x^2},{dc[1],dc[2]});

> c[1]:=int(-(3*x^2+2*ln(x))/x^2,x);c[2]:=int((3*x^2+2*ln(x))/x^3,x);

> c[1]*y[1]+c[2]*y[2];

> simplify(%);

So one particular solution is . To check that this is indeed a particular solution we plug it into the original equation:

> x^2*diff(-3*x^2+ln(x)+3/2+3*ln(x)*x^2,x,x)-2*x*diff(-3*x^2+ln(x)+3/2+3*ln(x)*x^2,x)+2*(-3*x^2+ln(x)+3/2+3*ln(x)*x^2)+2*(-3*x^2+ln(x)+3/2+3*ln(x)*x^2);

> simplify(%);

The general solution is .

Example 6: Solve .

Solution: First we find by solving the equation .

> solve(r*(r-1)+r+4=0,r);

Therefore the solutions to the homogenous part are eight=32 alt="[Maple Math]" align=middle> and . To find one particular solution of the original equation we use the method of variation as follows:

> y[1]:=cos(2*ln(x));y[2]:=sin(2*ln(x));

> dy[1]:=diff(y[1],x);dy[2]:=diff(y[2],x);

> solve({y[1]*dc[1]+y[2]*dc[2]=0,dy[1]*dc[1]+dy[2]*dc[2]=sin(ln(x))/x^2},{dc[1],dc[2]});

> c[1]:=int(-sin(ln(x))^2*cos(ln(x))/x,x);c[2]:=int(-1/2*(-1+2*sin(ln(x))^2)*sin(ln(x))/x,x);

> c[1]*y[1]+c[2]*y[2];

> simplify(%,{sin(ln(x))^2=1-cos(ln(x))^2});

So one particular solution is . To check that this is indeed a particular solution we plug it into the original equation:

> x^2*diff(1/3*sin(ln(x)),x,x)+x*diff(1/3*sin(ln(x)),x)+4*(1/3*sin(ln(x)));

> simplify(%);

The general solution is 000>The general solution is [Maple Math] .

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