problem
Verify that y_{1},y_{2} are solutions of the differential equation. Then find a particular solution of the form y=c_{1}y_{1}+c_{2}y_{2} that satisfies the initial conditions. y^{\prime \prime }-10y^{\prime }+25y=0 with y_{1}=e^{5x},y_{2}=xe^{5x} and y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =13
solution
To verify that y_{1} or y_{2} is solution to the ODE, we plug it into the ODE and see if it gives zero, which is what the RHS is. Since y_{1}^{\prime }=5e^{5x},y_{1}^{\prime \prime }=25e^{5x}, then substituting this into the ODE gives
\begin{align*} y_{1}^{\prime \prime }-10y_{1}^{\prime }+25y_{1} & =0\\ 25e^{5x}-10\left ( 5e^{5x}\right ) +25\left ( e^{5x}\right ) & =0\\ 25e^{5x}-50e^{5x}+25e^{5x} & =0\\ 0 & =0 \end{align*}
Hence verified. Now we do the same for y_{2}. Since y_{2}^{\prime }=e^{5x}+5xe^{5x},y_{2}^{\prime \prime }=5e^{5x}+5e^{5x}+25xe^{5x}, then substituting this into the ODE gives
\begin{align*} y_{2}^{\prime \prime }-10y_{2}^{\prime }+25y_{2} & =0\\ \left ( 5e^{5x}+5e^{5x}+25xe^{5x}\right ) -10\left ( e^{5x}+5xe^{5x}\right ) +25\left ( xe^{5x}\right ) & =0\\ 5e^{5x}+5e^{5x}+25xe^{5x}-10e^{5x}-50xe^{5x}+25xe^{5x} & =0\\ 25xe^{5x}-50xe^{5x}+25xe^{5x} & =0\\ 0 & =0 \end{align*}
Hence verified. Therefore the general solution is
y\left ( x\right ) =c_{1}y_{1}\left ( x\right ) +c_{2}y\left ( x\right )
Where the constants are found from initial conditions. Using the first initial condition gives
\begin{align*} y\left ( 0\right ) & =3\\ c_{1}y_{1}\left ( 0\right ) +c_{2}y_{2}\left ( 0\right ) & =3\\ c_{1}\left ( e^{5x}\right ) _{x=0}+c_{2}\left ( xe^{5x}\right ) _{x=0} & =3\\ c_{1} & =3 \end{align*}
Hence the solution becomes \begin{align*} y\left ( x\right ) & =3y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \\ y^{\prime } & =3y_{1}^{\prime }+c_{2}y_{2}^{\prime }\\ & =3\left ( 5e^{5x}\right ) +c_{2}\left ( e^{5x}+5xe^{5x}\right ) \end{align*}
Applying the second boundary conditions gives
\begin{align*} y^{\prime }\left ( 0\right ) & =13\\ 3\left ( 5e^{5x}\right ) _{x=0}+c_{2}\left ( e^{5x}+5xe^{5x}\right ) _{x=0} & =13\\ 3\left ( 5\right ) +c_{2} & =13\\ c_{2} & =13-15\\ & =-2 \end{align*}
Therefore the particular solution is
\begin{align*} y\left ( x\right ) & =c_{1}y_{1}\left ( x\right ) +c_{2}y\left ( x\right ) \\ & =3y_{1}\left ( x\right ) -2y\left ( x\right ) \\ & =3e^{5x}-2xe^{5x}\\ & =e^{5x}\left ( 3-2x\right ) \end{align*}
problem Show that y_{1}=1,y_{2}=\sqrt{x} are solutions to yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}=0 but that their sum y=y_{1}+y_{2} is not a solution
solution To show that y_{1} and y_{2} are solution to the ODE, we plug them into the ODE and see if the result is the same as the RHS. Since y_{1}=1 then y_{1}^{\prime }=0,y_{1}^{\prime \prime }=0. Then ODE becomes
\begin{align*} y_{1}y_{1}^{\prime \prime }+\left ( y_{1}^{\prime }\right ) ^{2} & =0\\ 1\left ( 0\right ) +0 & =0\\ 0 & =0 \end{align*}
Hence verified. For y_{2}, we have y_{2}^{\prime }=\frac{1}{2x^{\frac{1}{2}}},y_{2}^{\prime \prime }=\frac{-1}{4}\frac{1}{x^{\frac{3}{2}}}. Hence the ODE becomes
\begin{align*} y_{2}y_{2}^{\prime \prime }+\left ( y_{2}^{\prime }\right ) ^{2} & =0\\ x^{\frac{1}{2}}\left ( \frac{-1}{4}\frac{1}{x^{\frac{3}{2}}}\right ) +\left ( \frac{1}{2x^{\frac{1}{2}}}\right ) ^{2} & =0\\ \left ( \frac{-1}{4}\frac{1}{x}\right ) +\left ( \frac{1}{4x}\right ) & =0\\ 0 & =0 \end{align*}
Hence verified. Now we plugin the sum into the ODE.
\begin{align*} \left ( y_{1}+y_{2}\right ) \left ( y_{1}+y_{2}\right ) ^{\prime \prime }+\left ( \left ( y_{1}+y_{2}\right ) ^{\prime }\right ) ^{2} & =0\\ \left ( y_{1}+y_{2}\right ) \left ( y_{1}^{\prime \prime }+y_{2}^{\prime \prime }\right ) +\left ( y_{1}^{\prime }+y_{2}^{\prime }\right ) ^{2} & =0\\ \left ( y_{1}y_{1}^{\prime \prime }+y_{1}y_{2}^{\prime \prime }\right ) +\left ( y_{2}y_{1}^{\prime \prime }+y_{2}y_{2}^{\prime \prime }\right ) +\left ( y_{1}^{\prime }\right ) ^{2}+\left ( y_{2}^{\prime }\right ) ^{2}+2y_{1}^{\prime }y_{2}^{\prime } & =0\\ y_{1}y_{1}^{\prime \prime }+y_{1}y_{2}^{\prime \prime }+y_{2}y_{1}^{\prime \prime }+y_{2}y_{2}^{\prime \prime }+\left ( y_{1}^{\prime }\right ) ^{2}+\left ( y_{2}^{\prime }\right ) ^{2}+2y_{1}^{\prime }y_{2}^{\prime } & =0 \end{align*}
But we found that y_{1}y_{1}^{\prime \prime }+\left ( y_{1}^{\prime }\right ) ^{2}=0 and y_{2}y_{2}^{\prime \prime }+\left ( y_{2}^{\prime }\right ) ^{2}=0 from earlier. Using these into the LHS of the above simplifies it to
y_{1}y_{2}^{\prime \prime }+y_{2}y_{1}^{\prime \prime }+2y_{1}^{\prime }y_{2}^{\prime }=0
But y_{2}^{\prime \prime }=\frac{-1}{4}\frac{1}{x^{\frac{3}{2}}},y_{1}^{\prime \prime }=0,y_{1}^{\prime }=0,y_{1}=1, then the above becomes
\frac{-1}{4}\frac{1}{x^{\frac{3}{2}}}=0
We see that the LHS is not zero. Hence y_{1}+y_{2} is not a solution to the ODE.
problem Determine whether the pairs of functions are linearly independent or not on the real line. f\left ( x\right ) =\sin ^{2}x,g\left ( x\right ) =1-\cos 2x
solution The two functions are L.I. if c_{1}f\left ( x\right ) +c_{1}g\left ( x\right ) =0 for each x, only when c_{1}=c_{2}=0. Or stated differently, two functions are L.D. if there exist c_{1},c_{2} not all zero, such that c_{1}f\left ( x\right ) +c_{1}g\left ( x\right ) =0 for each x. To show this, we set up the Wronskian W and see if it is zero or not. If W=0 then this mean that the functions are L.D.
\begin{align*} W & =\begin{vmatrix} f\left ( x\right ) & g\left ( x\right ) \\ f^{\prime }\left ( x\right ) & g^{\prime }\left ( x\right ) \end{vmatrix} =\\ & \begin{vmatrix} \sin ^{2}x & 1-\cos 2x\\ 2\sin x\cos x & 2\sin 2x \end{vmatrix} \\ & =2\sin ^{2}x\sin 2x-\left ( 1-\cos 2x\right ) \left ( 2\sin x\cos x\right ) \\ & =2\sin ^{2}x\sin 2x-2\sin x\cos x+2\cos 2x\sin x\cos x \end{align*}
The RHS of the above simplifies to 0.
W=0
Therefore, the functions are linearly dependent.
problem Determine whether the pairs of functions are linearly independent or not on the real line. f\left ( x\right ) =2\cos x+3\sin x,g\left ( x\right ) =3\cos x-2\sin x
solution To show this, we set up the Wronskian W and see if it is zero or not. If W=0 then this mean that the functions are L.D.
\begin{align*} W & =\begin{vmatrix} f\left ( x\right ) & g\left ( x\right ) \\ f^{\prime }\left ( x\right ) & g^{\prime }\left ( x\right ) \end{vmatrix} =\\ & \begin{vmatrix} 2\cos x+3\sin x & 3\cos x-2\sin x\\ -2\sin x+3\cos x & -3\sin x-2\cos x \end{vmatrix} \\ & =\left ( 2\cos x+3\sin x\right ) \left ( -3\sin x-2\cos x\right ) -\left ( 3\cos x-2\sin x\right ) \left ( -2\sin x+3\cos x\right ) \\ & =-13\cos ^{2}x-13\sin ^{2}x\\ & =-13\left ( \cos ^{2}x+\sin ^{2}x\right ) \\ & =-13 \end{align*}
Since W\neq 0 then the functions are Linearly independent.
problem Ley y_{p} be a particular solution of the nonhomogeneous equation y^{\prime \prime }+py^{\prime }+qy=f\left ( x\right ) and let y_{h} be the homogenous solution. Show that y=y_{h}+y_{p} is a solution of the given ODE.
solution since y_{h} satisfies the homogenous ODE then we can write
\begin{equation} y_{h}^{\prime \prime }+py_{h}^{\prime }+qy_{h}=0 \tag{1} \end{equation}
And since y_{p} satisfies the nonhomogeneous ODE then we can write
\begin{equation} y_{p}^{\prime \prime }+py_{p}^{\prime }+qy_{p}=f\left ( x\right ) \tag{2} \end{equation}
Adding (1)+(2) gives
\left ( y_{p}^{\prime \prime }+y_{h}^{\prime \prime }\right ) +p\left ( y_{p}^{\prime }+y_{p}^{\prime }\right ) +q\left ( y_{p}+y_{h}\right ) =f\left ( x\right )
But due to linearity of differentiation, then the above can be written as
\left ( y_{p}+y_{h}\right ) ^{\prime \prime }+p\left ( y_{p}+y_{p}\right ) ^{\prime }+q\left ( y_{p}+y_{h}\right ) =f\left ( x\right )
Let Y=y_{p}+y_{h} then
Y^{\prime \prime }+pY^{\prime }+qY=f\left ( x\right )
Therefore we showed that Y=y_{p}+y_{h} satisfies the original ODE, hence it is a solution. QED
problem Show that y_{1}=\sin x^{2} and y=\cos x^{2} are L.I. functions, but their Wronskian vanishes are x=0. Why does this implies that there is no differential equation of the form y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 with both p,q continuous everywhere, having both y_{1},y_{2} are solutions?
solution
\begin{align*} W & =\begin{vmatrix} y_{1} & y_{2}\\ y_{2}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\\ & \begin{vmatrix} \sin x^{2} & \cos x^{2}\\ \left ( 2x\right ) \cos x^{2} & -\left ( 2x\right ) \sin x^{2}\end{vmatrix} \\ & =-2x\sin x^{2}\sin x^{2}-2x\cos ^{2}\cos x^{2}\\ & =-2x\left ( \left ( \sin x^{2}\right ) ^{2}+\left ( \cos x^{2}\right ) ^{2}\right ) \\ & =-2x \end{align*}
The Wronskian is zero at x=0 but not zero at other points. It is only when W=0 everywhere, we say that y_{1},y_{2} are L.D. We can have L.I. functions, but also have W\left ( x_{0}\right ) =0 at some x_{0} as in this problem. What this mean, is that x=0 can not be in the domain of the solution for y_{1},y_{2} to be solutions to the ODE. Hence, since the domain of the solution is everywhere, this means x=0 is part of the domain, then we conclude that y_{1},y_{2} can not be both solutions, since they are L.I. at x=0.
problem Let y_{1},y_{2} be two solutions of A\left ( x\right ) y^{\prime \prime }+B\left ( x\right ) y^{\prime }+C\left ( x\right ) y=0 on open interval I where A,B,C are continuous and A\left ( x\right ) is never zero. (a) Let W=W\left ( y_{1},y_{2}\right ) . Show that A\left ( x\right ) \frac{dW}{dx}=y_{1}\left ( Ay_{2}^{\prime \prime }\right ) -y_{2}\left ( Ay_{1}^{\prime \prime }\right ) then substitute for Ay_{2}^{\prime \prime } and Ay_{1}^{\prime \prime } from the original ODE to show that A\left ( x\right ) \frac{dW}{dx}=-B\left ( x\right ) W\left ( x\right ) (b) Solve this first order ODE equation to deduce Abel’s formula W\left ( x\right ) =k\exp \left ( -\int \frac{B\left ( x\right ) }{A\left ( x\right ) }dx\right ) where k is constant. (c) Why does Abel’s formula imply that the Wronskian W\left ( y_{1},y_{2}\right ) is either zero everywhere or non-zero everywhere (as stated in theorem 3)?
solution
By definition
W\left ( x\right ) =y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime }
Hence
\begin{align*} \frac{dW}{dx} & =y_{1}^{\prime }y_{2}^{\prime }+y_{1}y_{2}^{\prime \prime }-y_{2}^{\prime }y_{1}^{\prime }-y_{2}y_{1}^{\prime \prime }\\ & =y_{1}y_{2}^{\prime \prime }-y_{2}y_{1}^{\prime \prime } \end{align*}
Therefore
\begin{align} A\left ( x\right ) \frac{dW}{dx} & =A\left ( x\right ) \left ( y_{1}y_{2}^{\prime \prime }-y_{2}y_{1}^{\prime \prime }\right ) \nonumber \\ & =y_{1}\left ( A\left ( x\right ) y_{2}^{\prime \prime }\right ) -y_{2}\left ( A\left ( x\right ) y_{1}^{\prime \prime }\right ) \tag{1} \end{align}
But from original ODE, A\left ( x\right ) y_{1}^{\prime \prime }+B\left ( x\right ) y_{1}^{\prime }+C\left ( x\right ) y_{1}=0, therefore \begin{equation} A\left ( x\right ) y_{1}^{\prime \prime }=-B\left ( x\right ) y_{1}^{\prime }-C\left ( x\right ) y_{1} \tag{2} \end{equation}
Substituting (2,3) into (1) gives
\begin{align} A\left ( x\right ) \frac{dW}{dx} & =y_{1}\left ( -B\left ( x\right ) y_{2}^{\prime }-C\left ( x\right ) y_{2}\right ) -y_{2}\left ( -B\left ( x\right ) y_{1}^{\prime }-C\left ( x\right ) y_{1}\right ) \nonumber \\ & =-B\left ( x\right ) y_{1}y_{2}^{\prime }-C\left ( x\right ) y_{1}y_{2}+B\left ( x\right ) y_{2}y_{1}^{\prime }+C\left ( x\right ) y_{2}y_{1}\nonumber \\ & =-B\left ( x\right ) y_{1}y_{2}^{\prime }+B\left ( x\right ) y_{2}y_{1}^{\prime }\nonumber \\ & =-B\left ( x\right ) \left ( y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime }\right ) \nonumber \\ & =-B\left ( x\right ) W\left ( x\right ) \tag{4} \end{align}
QED.
Solving (4).
\begin{align*} A\left ( x\right ) \frac{dW}{dx}+B\left ( x\right ) W\left ( x\right ) & =0\\ \frac{dW}{dx}+\frac{B\left ( x\right ) }{A\left ( x\right ) }W\left ( x\right ) & =0 \end{align*}
Integrating factor is \mu =e^{\int \frac{B\left ( x\right ) }{A\left ( x\right ) }dx}, hence the above becomes
\frac{d}{dx}\left ( \mu W\left ( x\right ) \right ) =0
Integrating gives
\begin{align*} \mu W\left ( x\right ) & =k\\ W\left ( x\right ) & =ke^{-\int \frac{B\left ( x\right ) }{A\left ( x\right ) }dx} \end{align*}
Since an exponential function is never zero (for bounded \frac{B\left ( x\right ) }{A\left ( x\right ) }), then W\left ( x\right ) =ke^{\left ( \cdot \right ) } can only be zero if k=0. This makes W=0 everywhere when k=0. But if k\neq 0, then W\neq 0 everywhere. So W can only be zero everywhere, or not zero everywhere.
problem Apply theorem 5 and 6 to find general solutions of the differential equation y^{\prime \prime }+2y^{\prime }-15y=0
solution The characteristic equation is r^{2}+2r-15=0, and the roots are \begin{align*} r_{1} & =3\\ r_{2} & =-5 \end{align*}
Therefore the solution is
\begin{align*} y\left ( x\right ) & =c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}\\ & =c_{1}e^{3x}+c_{2}e^{-5x} \end{align*}
problem Apply theorem 5 and 6 to find general solutions of the differential equation 35y^{\prime \prime }-y^{\prime }-12y=0
solution The characteristic equation is 35r^{2}-r-12=0, and the roots are \begin{align*} r_{1} & =\frac{3}{5}\\ r_{2} & =-\frac{4}{7} \end{align*}
Therefore the solution is
\begin{align*} y\left ( x\right ) & =c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}\\ & =c_{1}e^{\frac{3}{5}x}+c_{2}e^{-\frac{4}{7}x} \end{align*}
problem Problem gives a general solution y\left ( x\right ) of a homogeneous second order ODE ay^{\prime \prime }+by^{\prime }+cy=0 with constant coefficients. Find such an equation y\left ( x\right ) =e^{x}\left ( c_{1}e^{x\sqrt{2}}+c_{2}e^{-x\sqrt{2}}\right )
solution We compare the above solution to the general form of the solution given by
\begin{align*} y & =c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}\\ & =c_{1}e^{x\left ( 1+\sqrt{2}\right ) }+c_{2}e^{x\left ( 1-\sqrt{2}\right ) } \end{align*}
We see that \begin{align*} r_{1} & =1+\sqrt{2}\\ r_{2} & =1-\sqrt{2} \end{align*}
This implies that the characteristic equation is
\begin{align*} \left ( r-r_{1}\right ) \left ( r-r_{2}\right ) & =0\\ \left ( r-\left ( 1+\sqrt{2}\right ) \right ) \left ( r-\left ( 1-\sqrt{2}\right ) \right ) & =0\\ r^{2}-2r-1 & =0 \end{align*}
Therefore the ODE is
y''-2y'-y=0
Where a=1,b=-2,c=-1.