In Problems 7 through 12, use the Wronskian to prove that the given functions are linearly independent on the indicated interval.
\(f\left ( x\right ) =e^{x},g\left ( x\right ) =\cos x,h\left ( x\right ) =\sin x\)
On the real line.
Solution
\[ W\left ( x\right ) =\begin {bmatrix} f\left ( x\right ) & g\left ( x\right ) & h\left ( x\right ) \\ f^{\prime }\left ( x\right ) & g^{\prime }\left ( x\right ) & h^{\prime }\left ( x\right ) \\ f^{\prime \prime }\left ( x\right ) & g^{\prime \prime }\left ( x\right ) & h^{\prime \prime }\left ( x\right ) \end {bmatrix} \] Hence\[ W\left ( x\right ) =\begin {bmatrix} e^{x} & \cos x & \sin x\\ e^{x} & -\sin x & \cos x\\ e^{x} & -\cos x & -\sin x \end {bmatrix} \] The determinant is, expanding along first row is\begin {align*} \left \vert W\left ( x\right ) \right \vert & =e^{x}\begin {vmatrix} -\sin x & \cos x\\ -\cos x & -\sin x \end {vmatrix} -\cos x\begin {vmatrix} e^{x} & \cos x\\ e^{x} & -\sin x \end {vmatrix} +\sin x\begin {vmatrix} e^{x} & -\sin x\\ e^{x} & -\cos x \end {vmatrix} \\ & =e^{x}\left ( \sin ^{2}x+\cos ^{2}x\right ) -\cos x\left ( -e^{x}\sin x-e^{x}\cos x\right ) +\sin x\left ( -e^{x}\cos x+e^{x}\sin x\right ) \end {align*}
But \(\sin ^{2}x+\cos ^{2}x=1\) and the above simplifies to\begin {align*} \left \vert W\left ( x\right ) \right \vert & =e^{x}-\left ( -e^{x}\sin x\cos x-e^{x}\cos ^{2}x\right ) +\left ( -e^{x}\cos x\sin x+e^{x}\sin ^{2}x\right ) \\ & =e^{x}+e^{x}\sin x\cos x+e^{x}\cos ^{2}x-e^{x}\cos x\sin x+e^{x}\sin ^{2}x\\ & =e^{x}+e^{x}\cos ^{2}x+e^{x}\sin ^{2}x\\ & =e^{x}+e^{x}\left ( \sin ^{2}x+\cos ^{2}x\right ) \\ & =2e^{x} \end {align*}
And since \(e^{x}\) is never zero on the real line, then \(\left \vert W\left ( x\right ) \right \vert \neq 0\) Hence functions are linearly independent.
In Problems 13 through 20, a third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions.\begin {align*} y^{\prime \prime \prime }-5y^{\prime \prime }+8y^{\prime }-4y & =0\\ y_{1} & =e^{x}\\ y_{2} & =e^{2x}\\ y_{3} & =xe^{2x} \end {align*}
I.C. are
\(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =4,y^{\prime \prime }\left ( 0\right ) =0\)
Solution
The general solution is\begin {align} y\left ( x\right ) & =c_{1}y_{2}+c_{2}y_{2}+c_{3}y_{3}\nonumber \\ & =c_{1}e^{x}+c_{2}e^{2x}+c_{3}xe^{2x} \tag {1} \end {align}
At \(y\left ( 0\right ) =0\) the above becomes\begin {equation} 1=c_{1}+c_{2} \tag {2} \end {equation} Taking derivative of (1) gives\[ y^{\prime }\left ( x\right ) =c_{1}e^{x}+2c_{2}e^{2x}+c_{3}\left ( e^{2x}+2xe^{2x}\right ) \] At \(y^{\prime }\left ( 0\right ) =4\) the above becomes\begin {equation} 4=c_{1}+2c_{2}+c_{3} \tag {3} \end {equation} Taking derivative of \(y^{\prime \prime }\left ( x\right ) \) gives\begin {align*} y^{\prime \prime }\left ( x\right ) & =c_{1}e^{x}+4c_{2}e^{2x}+c_{3}\left ( 2e^{2x}+2\left ( e^{2x}+2xe^{2x}\right ) \right ) \\ & =c_{1}e^{x}+4c_{2}e^{2x}+c_{3}\left ( 2e^{2x}+2e^{2x}+4xe^{2x}\right ) \end {align*}
At \(y^{\prime \prime }\left ( 0\right ) =0\) the above becomes\begin {equation} 0=c_{1}+4c_{2}+4c_{3} \tag {4} \end {equation} Equations (2,3,4) are now solved for \(c_{1},c_{2},c_{3}\)\[\begin {bmatrix} 1 & 1 & 0\\ 1 & 2 & 1\\ 1 & 4 & 4 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} 1\\ 4\\ 0 \end {bmatrix} \] Augmented matrix\[\begin {bmatrix} 1 & 1 & 0 & 1\\ 1 & 2 & 1 & 4\\ 1 & 4 & 4 & 0 \end {bmatrix} \] \(R_{2}\rightarrow R_{2}-R_{1}\)\[\begin {bmatrix} 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 3\\ 1 & 4 & 4 & 0 \end {bmatrix} \] \(R_{3}\rightarrow R_{3}-R_{1}\)\[\begin {bmatrix} 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 3\\ 0 & 3 & 4 & -1 \end {bmatrix} \] \(R_{3}\rightarrow R_{3}-3R_{2}\)\[\begin {bmatrix} 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 3\\ 0 & 0 & 1 & -10 \end {bmatrix} \] The above is Echelon form. Hence the system becomes\[\begin {bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} 1\\ 3\\ -10 \end {bmatrix} \] From last row, \(c_{3}=-10\). From second row \(c_{2}+c_{3}=3\) or \(c_{2}=13\). From first row \(c_{1}+c_{2}=1\). Hence \(c_{1}=-12\). Therefore\[\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} -12\\ 13\\ -10 \end {bmatrix} \] Substituting these values back in general solution (1) gives the solution that satisfies these initial conditions as\begin {align*} y\left ( x\right ) & =c_{1}e^{x}+c_{2}e^{2x}+c_{3}xe^{2x}\\ & =-12e^{x}+13e^{2x}-10xe^{2x} \end {align*}
In Problems 13 through 20, a third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions.\begin {align*} x^{3}y^{\prime \prime \prime }-3x^{2}y^{\prime \prime }+6xy^{\prime }-6y & =0\\ y_{1} & =x\\ y_{2} & =x^{2}\\ y_{3} & =x^{3} \end {align*}
I.C. are
\(y\left ( 1\right ) =6,y^{\prime }\left ( 1\right ) =14,y^{\prime \prime }\left ( 1\right ) =22\)
Solution
The general solution is\begin {align} y\left ( x\right ) & =c_{1}y_{2}+c_{2}y_{2}+c_{3}y_{3}\nonumber \\ & =c_{1}x+c_{2}x^{2}+c_{3}x^{3}\tag {1} \end {align}
At \(y\left ( 1\right ) =0\) the above becomes\begin {equation} 6=c_{1}+c_{2}+c_{3}\tag {2} \end {equation} Taking derivative of (1) gives\[ y^{\prime }\left ( x\right ) =c_{1}+2c_{2}x+3c_{3}x^{2}\] At \(y^{\prime }\left ( 1\right ) =14\) the above becomes\begin {equation} 14=c_{1}+2c_{2}+3c_{3}\tag {3} \end {equation} Taking derivative of \(y^{\prime }\left ( x\right ) \) gives\[ y^{\prime \prime }\left ( x\right ) =2c_{2}+6c_{3}x \] At \(y^{\prime \prime }\left ( 1\right ) =22\) the above becomes\begin {equation} 22=2c_{2}+6c_{3}\tag {4} \end {equation} Equations (2,3,4) are now solved for \(c_{1},c_{2},c_{3}\)\[\begin {bmatrix} 1 & 1 & 1\\ 1 & 2 & 3\\ 0 & 2 & 6 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} 6\\ 14\\ 22 \end {bmatrix} \] Augmented matrix\[\begin {bmatrix} 1 & 1 & 1 & 6\\ 1 & 2 & 3 & 14\\ 0 & 2 & 6 & 22 \end {bmatrix} \] \(R_{2}\rightarrow R_{2}-R_{1}\)\[\begin {bmatrix} 1 & 1 & 1 & 6\\ 0 & 1 & 2 & 8\\ 0 & 2 & 6 & 22 \end {bmatrix} \] \(R_{3}\rightarrow R_{3}-2R_{2}\)\[\begin {bmatrix} 1 & 1 & 1 & 6\\ 0 & 1 & 2 & 8\\ 0 & 0 & 2 & 6 \end {bmatrix} \] The above is Echelon form. Hence the system becomes\[\begin {bmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} 6\\ 8\\ 6 \end {bmatrix} \] From last row, \(2c_{3}=6\) or \(c_{3}=3\). From second row \(c_{2}+2c_{3}=8\) or \(c_{2}=8-2\left ( 3\right ) =2\). From first row \(c_{1}+c_{2}+c_{3}=6\). Hence \(c_{1}=6-2-3=1\). Therefore\[\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \] Substituting these values back in general solution (1) gives the solution that satisfies these initial conditions as\begin {align*} y\left ( x\right ) & =c_{1}x+c_{2}x^{2}+c_{3}x^{3}\\ & =x+2x^{2}+3x^{3} \end {align*}
In Problems 21 through 24, a nonhomogeneous differential equation, a complementary solution \(y_{c}\), and a particular solution
\(y_{p}\) are given. Find a solution satisfying the given initial conditions.\begin {align*} y^{\prime \prime }-2y^{\prime }+2y & =2x\\ y_{c} & =c_{1}e^{x}\cos x+c_{2}e^{x}\sin x\\ y_{p} & =x+1 \end {align*}
I.C. are
\(y\left ( 0\right ) =4,y^{\prime }\left ( 0\right ) =8\)
Solution
The general solution is\begin {align} y\left ( x\right ) & =y_{c}+y_{p}\nonumber \\ & =c_{1}e^{x}\cos x+c_{2}e^{x}\sin x+x+1 \tag {1} \end {align}
At \(y\left ( 0\right ) =4\) the above becomes (using \(e^{0}=1,\cos 0=1,\sin 0=0\))\begin {equation} 4=c_{1}+1 \tag {2} \end {equation} Taking derivative of (1) gives\[ y^{\prime }\left ( x\right ) =c_{1}\left ( e^{x}\cos x-e^{x}\sin x\right ) +c_{2}e^{x}\cos x+1 \] At \(y^{\prime }\left ( 0\right ) =8\) the above becomes\begin {align} 8 & =c_{1}\left ( 1-0\right ) +c_{2}+1\nonumber \\ 8 & =c_{1}+c_{2}+1 \tag {3} \end {align}
We have two equations (2,3) to solve for \(c_{1},c_{2}\). From (3) we see that \(c_{1}=3\). Hence from (3) \(8=3+c_{2}+1\) or \(c_{2}=4\). Therefore the solution in (1) becomes\begin {align*} y\left ( x\right ) & =c_{1}e^{x}\cos x+c_{2}e^{x}\sin x+x+1\\ & =3e^{x}\cos x+4e^{x}\sin x+x+1\\ & =e^{x}\left ( 3\cos x+4\sin x\right ) +x+1 \end {align*}
Find the general solutions of the differential equations in Problems 1 through 20.\[ y^{\prime \prime }-6y^{\prime }+13y=0 \] Solution This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ Ay^{\prime \prime }(x)+By^{\prime }(x)+Cy(x)=0 \] Where here we see that \(A=1,B=-6,C=13\).
Let the solution be \(y\left ( x\right ) =e^{\lambda x}\). Substituting this into the ODE gives \begin {equation} \lambda ^{2}e^{\lambda x}-6\lambda e^{\lambda x}+13e^{\lambda x}=0 \tag {1} \end {equation}
Since \(e^{\lambda x}\neq 0\), then dividing Eq. (1) throughout by\(e^{\lambda x}\) results in \begin {equation} \lambda ^{2}-6\lambda +13=0 \tag {2} \end {equation}
Eq. (2) is the characteristic equation of the ODE. We need to determine its roots to find the general solution. Using the quadratic formula \[ \lambda _{1,2}=\frac {-B}{2A}\pm \frac {1}{2A}\sqrt {B^{2}-4AC}\] Substituting \(A=1,B=-6,C=13\) into the above gives \begin {align*} \lambda _{1,2} & =\frac {6}{(2)\left ( 1\right ) }\pm \frac {1}{(2)\left ( 1\right ) }\sqrt {-6^{2}-(4)\left ( 1\right ) \left ( 13\right ) }\\ & =3\pm 2i \end {align*}
Hence \begin {align*} \lambda _{1} & =3+2i\\ \lambda _{2} & =3-2i \end {align*}
Since roots are complex conjugate of each others, then let the roots be \[ \lambda _{1,2}=\alpha \pm i\beta \] Where \(\alpha =3\) and \(\beta =2\). Therefore the final solution, when using Euler relation, can be written as \[ y\left ( x\right ) =e^{\alpha x}\left ( c_{1}\cos (\beta x)+c_{2}\sin (\beta x)\right ) \] Which becomes \[ y\left ( x\right ) =e^{3x}\left ( c_{1}\cos \left ( 2x\right ) +c_{2}\sin \left ( 2x\right ) \right ) \]
Find the general solutions of the differential equations in Problems 1 through 20.\[ y^{\left ( 4\right ) }\left ( x\right ) -8y^{\left ( 3\right ) }+16y^{\prime \prime }=0 \] Solution
We start by writing the characteristic equation of the ODE \[ \lambda ^{4}-8\lambda ^{3}+16\lambda ^{2}=0 \] We now solve for the roots of the above equation. Writing the above as\[ \lambda ^{2}\left ( \lambda ^{2}-8\lambda +16\right ) =0 \] We see that \(\lambda ^{2}=0\) gives \(\lambda =0\) with multiplicity \(2\,\). The equation \(\lambda ^{2}-8\lambda +16=0\) can be factored to \(\left ( \lambda -4\right ) \left ( \lambda -4\right ) =0\). Therefor \(\lambda =4\) with multiplicity \(2\).
Hence the roots are\begin {align*} \lambda _{1} & =0\\ \lambda _{2} & =0\\ \lambda _{3} & =4\\ \lambda _{4} & =4 \end {align*}
This table summarizes the result
| root | multiplicity | type of root |
| \(0\) | \(2\) | real root |
| \(4\) | \(2\) | real root |
For a real root \(\lambda \) with multiplicity one, we obtain a basis solution of the form \(e^{\lambda x}\) and real root \(\lambda \) with multiplicity two we obtain basis solutions \(\left \{ e^{\lambda x},xe^{\lambda x}\right \} \). Therefore the solution is\begin {align*} y\left ( x\right ) & =c_{2}e^{\lambda _{1}x}+c_{2}xe^{\lambda _{1}x}+c_{2}e^{\lambda _{3}x}+c_{2}xe^{\lambda _{3}x}\\ & =c_{2}+c_{2}x+c_{2}e^{4x}+c_{2}xe^{4x} \end {align*}
Find the general solutions of the differential equations in Problems 1 through 20.\[ y^{\left ( 4\right ) }\left ( x\right ) +3y^{\prime \prime }-4y=0 \] Solution
We start by writing the characteristic equation \[ \lambda ^{4}+3\lambda ^{2}-4=0 \] Let \[ z=\lambda ^{2}\] The characteristic becomes\[ z^{2}+3z-4=0 \] Factoring the above gives\[ \left ( z+4\right ) \left ( z-1\right ) =0 \] Hence \(z=-4,z=1\). When \(z=-4\), then \(\lambda =\pm \sqrt {-4}=\pm 2i\). And when \(z=1\), then \(\lambda =\pm \sqrt {1}=\pm 1\). Therefore the roots are
\begin {align*} \lambda _{1} & =1\\ \lambda _{2} & =-1\\ \lambda _{3} & =2i\\ \lambda _{4} & =-2i \end {align*}
This table summarizes the result
| root | multiplicity | type of root |
| \(-1\) | \(1\) | real root |
| \(1\) | \(1\) | real root |
| \(\pm 2 i\) | \(1\) | complex conjugate root |
For a real root \(\lambda \) with multiplicity one, we obtain a basis of the form \(c_{1}e^{\lambda x}\) and for a complex conjugate root of the form \(a\pm ib\) we obtain basis solution of the form \(e^{ax}\left ( c_{1}\cos \left ( bx\right ) +c_{2}\sin \left ( bx\right ) \right ) \). Therefore the final solution, using \(a=0,b=2\) is\[ y\left ( x\right ) =c_{1}e^{-x}+c_{2}e^{x}+c_{3}\cos \left ( 2x\right ) +c_{4}\sin \left ( 2x\right ) \]
Find the general solutions of the differential equations in Problems 1 through 20.\[ y^{\left ( 4\right ) }\left ( x\right ) =16y \] Solution
We start by writing the characteristic equation \[ \lambda ^{4}=16 \] Let\[ z=\lambda ^{2}\] The characteristic becomes\[ z^{2}=16 \] Hence \(z=\pm 4\). When \(z=4\) then \(\lambda =\pm \sqrt {4}=\pm 2\). And when \(z=-4\) then \(\lambda =\pm \sqrt {-4}=\pm 2i\). Hence the roots are\begin {align*} \lambda _{1} & =2\\ \lambda _{2} & =-2\\ \lambda _{3} & =2i\\ \lambda _{4} & =-2i \end {align*}
This table summarizes the result
| root | multiplicity | type of root |
| \(-2\) | \(1\) | real root |
| \(2\) | \(1\) | real root |
| \(\pm 2 i\) | \(1\) | complex conjugate root |
As in the earlier problem, we now can write the general solution as\[ y\left ( x\right ) =e^{-2x}c_{1}+c_{2}e^{2x}+c_{3}\cos \left ( 2x\right ) +c_{4}\sin \left ( 2x\right ) \]
Find the general solutions of the differential equations in Problems 1 through 20.\[ y^{\left ( 7\right ) }\left ( x\right ) -2y^{\left ( 6\right ) }+9y^{\left ( 5\right ) }-16y^{\left ( 4\right ) }+24y^{\left ( 3\right ) }-32y^{\prime \prime }+16y^{\prime }=0 \] Solution
The characteristic equation is
\begin {align*} \,r^{7}-2r^{6}+9r^{5}-16r^{4}+24r^{3}-32r^{2}+16r & =0\\ r\left ( r^{6}-2r^{5}+9r^{4}-16r^{3}+24r^{2}-32r+16\right ) & =0 \end {align*}
Hence one root is \(r=0\). And now we need to solve\[ r^{6}-2r^{5}+9r^{4}-16r^{3}+24r^{2}-32r+16=0 \]
Substituting \(r=1\) in the above gives\begin {align*} 1-2+9-16+24-32+16 & =0\\ 0 & =0 \end {align*}
Therefore \(\left ( r-1\right ) \) is a factor. Doing long division (do not know how type polynomial division in Latex, please see scanned hand solution in appendix of this problem).\[ \frac {r^{6}-2r^{5}+9r^{4}-16r^{3}+24r^{2}-32r+16}{\left ( r-1\right ) }=r^{5}-r^{4}+8r^{3}-8r^{2}+16r-16 \] Hence \[ r^{6}-2r^{5}+9r^{4}-16r^{3}+24r^{2}-32r+16=\left ( r-1\right ) \left ( r^{5}-r^{4}+8r^{3}-8r^{2}+16r-16\right ) \] Substituting \(r=1\) in \(\left ( r^{5}-r^{4}+8r^{3}-8r^{2}+16r-18\right ) \) gives\[ r^{5}-r^{4}+8r^{3}-8r^{2}+16r-18\rightarrow 1-1+8-8+16-16=0 \] Hence \(\left ( r-1\right ) \) is a factor of \(\left ( r^{5}-r^{4}+8r^{3}-8r^{2}+16r-16\right ) \). Therefore we now need to do long division\[ \frac {r^{5}-r^{4}+8r^{3}-8r^{2}+16r-16}{\left ( r-1\right ) }=r^{4}+8r^{2}+16 \] Hence now we have \[ r^{6}-2r^{5}+9r^{4}-16r^{3}+24r^{2}-32r+16=\left ( r-1\right ) \left ( r-1\right ) \left ( r^{4}+8r^{2}+16\right ) \]
Looking at \(r^{4}+8r^{2}+16=0\), let \(z=r^{2}\). Therefore \(r^{4}+8r^{2}+16\) becomes \(z^{2}+8z+16=0\,\), This can be factored to \(\left ( z+4\right ) \left ( z+4\right ) =0\). Hence roots are \(z=-4\) which is double root.
Therefore when \(z=-4\) then \(r=\pm \sqrt {-4}=\pm 2i\) with multiplicity \(2\) since \(z=-4\) is double root. Therefore the final factorization is\[ r^{6}-2r^{5}+9r^{4}-16r^{3}+24r^{2}-32r+16=\left ( r-1\right ) \left ( r-1\right ) \left ( r-2i\right ) \left ( r+2i\right ) \left ( r-2i\right ) \left ( r+2i\right ) \]
This table summarizes the result
| root | multiplicity | type of root |
| \(0\) | \(1\) | real root |
| \(1\) | \(2\) | real root |
| \(\pm 2i\) | \(2\) | complex conjugate |
Now we are above to write down the general solution.\begin {align*} y\left ( x\right ) & =c_{1}e^{0x}+\left ( c_{2}e^{x}+c_{3}xe^{x}\right ) +\left ( c_{4}e^{2ix}+c_{5}xe^{2ix}\right ) +\left ( c_{6}e^{-2ix}+c_{7}xe^{-2ix}\right ) \\ & =c_{1}+\left ( c_{2}e^{x}+c_{3}xe^{x}\right ) +\left ( c_{4}e^{2ix}+c_{5}xe^{2ix}\right ) +\left ( c_{6}e^{-2ix}+c_{7}xe^{-2ix}\right ) \\ & =c_{1}+\left ( c_{2}e^{x}+c_{3}xe^{x}\right ) +\left ( c_{4}e^{2ix}+c_{6}e^{-2ix}\right ) +x\left ( c_{5}e^{2ix}+c_{7}e^{-2ix}\right ) \end {align*}
We see the above has \(7\) terms. But using Euler relation, we can write \(\left ( e^{2ix}+e^{-2ix}\right ) \) using trig functions. The above becomes
\[ y\left ( x\right ) =c_{1}+\left ( c_{2}e^{x}+c_{3}xe^{x}\right ) +\left ( c_{4}\cos 2x+c_{5}\sin 2x\right ) +x\left ( c_{6}\cos 2x+c_{7}\sin 2x\right ) \]
(constants of integrations kept the same as originally for simplicity, since it does not matter as these are found from initial conditions if given).
