ODE5.1

From Example Problems
Jump to: navigation, search


x^{2}y''+7xy'+8y=0\,

The adjoint of an equation in the form

a(x)y''+b(x)y'+c(x)y=0\, is

a(x)y''+[2a'(x)-b(x)]y'+[a''(x)-b'(x)+c(x)]y=0\,

So the adjoint for this DE is:

x^{2}y''-3xy'+3y=0\,

Any solution of the adjoint equation is an integrating factor for the original problem.

y=x\, is one solution. Multiply x\, through the original DE.

x^{3}y''+7x^{2}y'+8xy=0\,

This is an exact equation, since

{\frac  {d}{dx}}[a(x)y'+b(x)y]=a'(x)y'+a(x)y''+b'(x)y+b(x)y'\,
=a(x)y''+[a'(x)+b(x)]y'+b'(x)y\,

In the case of an exact equation the last relation equals 0, so

a(x)=x^{3},a'(x)=3x^{2}\,
b'(x)=8x,b(x)=4x^{2}\,

{\frac  {d}{dx}}[x^{3}y'+4x^{2}y]=0\,. Integrating,

x^{3}y'+4x^{2}y=c_{1}\,
y'+4x^{{-1}}y=c_{1}x^{{-3}}\,

Use the integrating factor e^{{\int 4x^{{-1}}dx}}=x^{4}\,.

x^{4}y'+4x^{3}y=c_{1}x\,
{\frac  {d}{dx}}[x^{4}y]=c_{1}x\,
x^{4}y={\frac  {c_{1}}{2}}x^{2}+c_{2}\,

Finally,

y={\frac  {c_{1}}{2}}x^{{-2}}+c_{2}x^{{-4}}\,




==Alternate Solution==

x^{2}y''+7xy'+8y=0\,

This equation is recognizable as an Euler equation since the exponent on the x is the same as the order of the derivative in each term. So try for a solution in the form y=x^{r}.

y=x^{r},\quad y'=rx^{{r-1}},\quad y''=r(r-1)x^{{r-2}}

Substituting these in the equation gives:

x^{2}r(r-1)x^{{(r-2)}}+7xrx^{{(r-1)}}+8x^{r}=0\,

x^{r}(r^{2}+6r+8)=0\,

(r+2)(r+4)=0,\quad r=-2,\ -4\,

The solution is therefore y=Ax^{{-2}}+Bx^{{-4}},\ x\neq 0\,.

Ordinary Differential Equations

Main Page