Problem: Use the Wronskian to prove that the given functions are linearly independent on the given interval. f\left ( x\right ) =x,g\left ( x\right ) =\cos \left ( \ln x\right ) ,h\left ( x\right ) =\sin \left ( \ln x\right ) for x>0
solution The Wronskian is
\begin{align*} W\left ( x\right ) & =\begin{vmatrix} f & g & h\\ f^{\prime } & g^{\prime } & h^{\prime }\\ f^{\prime \prime } & g^{\prime \prime } & h^{\prime \prime }\end{vmatrix} \\ & =\begin{vmatrix} x & \cos \left ( \ln x\right ) & \sin \left ( \ln x\right ) \\ 1 & -\sin \left ( \ln x\right ) \frac{1}{x} & \cos \left ( \ln x\right ) \frac{1}{x}\\ 0 & -\cos \left ( \ln x\right ) \frac{1}{x^{2}}+\sin \left ( \ln \right ) \frac{1}{x^{2}} & -\sin \left ( \ln x\right ) \frac{1}{x^{2}}-\cos \left ( \ln x\right ) \frac{1}{x^{2}}\end{vmatrix} \end{align*}
Expanding along the last row\begin{align*} W\left ( x\right ) & =W_{32}\left ( -1\right ) ^{3+2}A_{32}+W_{33}\left ( -1\right ) ^{3+3}A_{33}\\ & =-\left ( -\cos \left ( \ln x\right ) \frac{1}{x^{2}}+\sin \left ( \ln \right ) \frac{1}{x^{2}}\right ) \begin{vmatrix} x & \sin \left ( \ln x\right ) \\ 1 & \cos \left ( \ln x\right ) \frac{1}{x}\end{vmatrix} +\left ( -\sin \left ( \ln x\right ) \frac{1}{x^{2}}-\cos \left ( \ln x\right ) \frac{1}{x^{2}}\right ) \begin{vmatrix} x & \cos \left ( \ln x\right ) \\ 1 & -\sin \left ( \ln x\right ) \frac{1}{x}\end{vmatrix} \\ & =\left ( \cos \left ( \ln x\right ) \frac{1}{x^{2}}-\sin \left ( \ln \right ) \frac{1}{x^{2}}\right ) \left ( \cos \left ( \ln x\right ) -\sin \left ( \ln x\right ) \right ) +\left ( \sin \left ( \ln x\right ) \frac{1}{x^{2}}+\cos \left ( \ln x\right ) \frac{1}{x^{2}}\right ) \left ( \sin \left ( \ln x\right ) +\cos \left ( \ln x\right ) \right ) \end{align*}
Let \sin \left ( \ln \right ) \frac{1}{x^{2}}=A,\cos \left ( \ln x\right ) \frac{1}{x^{2}}=B, \cos \left ( \ln x\right ) =a,\sin \left ( \ln x\right ) =b then the above is\begin{align*} W\left ( x\right ) & =\left ( B-A\right ) \left ( a-b\right ) +\left ( A+B\right ) \left ( b+a\right ) \\ & =2Ab+2Ba \end{align*}
Transforming back\begin{align*} W\left ( x\right ) & =2\sin \left ( \ln \right ) \frac{1}{x^{2}}\sin \left ( \ln x\right ) +2\cos \left ( \ln x\right ) \frac{1}{x^{2}}\cos \left ( \ln x\right ) \\ & =2\sin ^{2}\left ( \ln \right ) \frac{1}{x^{2}}+2\cos ^{2}\left ( \ln x\right ) \frac{1}{x^{2}}\\ & =\frac{2}{x^{2}} \end{align*}
Hence, for x>0 the Wronskian is not zero. Therefore the functions are L.I.
Problem: A third order ODE is given, and three L.I. solutions are given. Find a particular solution satisfying the given initial conditions y^{\prime \prime \prime }-5y^{\prime \prime }+8y^{\prime }-4y=0 and y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =4,y^{\prime \prime }\left ( 0\right ) =0 and y_{1}=e^{x},y_{2}=e^{2x},y_{3}=xe^{2x}
solution The general solution is\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}\\ & =c_{1}e^{x}+c_{2}e^{2x}+c_{3}xe^{2x} \end{align*}
Hence y^{\prime }=c_{1}e^{x}+2c_{2}e^{2x}+c_{3}\left ( e^{2x}+2xe^{2x}\right ) And y^{\prime \prime }=c_{1}e^{x}+4c_{2}e^{2x}+c_{3}\left ( 2e^{2x}+2e^{2x}+4xe^{2x}\right ) From first initial condition we obtain\begin{equation} 1=c_{1}+c_{2} \tag{1} \end{equation} From second initial condition we obtain\begin{equation} 4=c_{1}+2c_{2}+c_{3} \tag{2} \end{equation} And from the third initial condition\begin{equation} 0=c_{1}+4c_{2}+4c_{3} \tag{3} \end{equation} We have three equations (1,2,3) to solve for c_{1},c_{2},c_{3}.\begin{pmatrix} 1 & 1 & 0\\ 1 & 2 & 1\\ 1 & 4 & 4 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 1\\ 4\\ 0 \end{pmatrix} Augmented matrix is\begin{pmatrix} 1 & 1 & 0 & 1\\ 1 & 2 & 1 & 4\\ 1 & 4 & 4 & 0 \end{pmatrix} \overset{R_{2}=R_{2}-R_{1}}{\underset{R_{3}=R_{3}-R_{1}}{\longmapsto }}\begin{pmatrix} 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 3\\ 0 & 3 & 4 & -1 \end{pmatrix} \overset{R_{3}=R_{3}-3R_{2}}{\longmapsto }\begin{pmatrix} 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 3\\ 0 & 0 & 1 & -10 \end{pmatrix} We see that \left \vert A\right \vert =1, since reduced matrix is upper diagonal matrix. Hence the solution is unique. From last row we obtain c_{3}=-10 and from second row c_{2}+c_{3}=3 or c_{2}=3+10=13 and from first row c_{1}+c_{2}=1 or c_{1}=1-13=-12, hence the particular solution is\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*}
Problem: A nonhomogeneous ODE, homogeneous solution y_{h} and particular solution y_{p} are given. Find solution that satisfy the initial conditions. y^{\prime \prime }-2y^{\prime }+2y=2x with y\left ( 0\right ) =4,y^{\prime }\left ( 0\right ) =8 and y_{h}=c_{1}e^{x}\cos x+c_{2}e^{x}\sin x and y_{p}=x+1
solution The general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{x}\cos x+c_{2}e^{x}\sin x+x+1 \end{align*}
Therefore y^{\prime }=c_{1}\left ( e^{x}\cos x-e^{x}\sin x\right ) +c_{2}\left ( e^{x}\sin x+e^{x}\cos x\right ) +1 First initial conditions gives\begin{align*} 4 & =c_{1}+1\\ c_{1} & =3 \end{align*}
Second initial conditions gives 8=c_{1}+c_{2}+1 Hence c_{2}=7-c_{1}=4. Therefore the general solution becomes\begin{align*} y & =3e^{x}\cos x+4e^{x}\sin x+x+1\\ & =e^{x}\left ( 3\cos x+4\sin x\right ) +x+1 \end{align*}
Problem: Show that 1,x,x^{2},\cdots ,x^{n} are L.I.
solution Using the Wronskian W\left ( x\right ) =\begin{vmatrix} 1 & x & x^{2} & x^{3} & \cdots & x^{n}\\ 0 & 1 & 2x & 3x^{2} & \cdots & nx^{n-1}\\ 0 & 0 & 2 & 6x & \cdots & n\left ( n-1\right ) x^{n-2}\\ \vdots & \vdots & 0 & 6 & \cdots & n\left ( n-1\right ) \left ( n-2\right ) x^{n-3}\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & & n! \end{vmatrix} Therefore, the resulting Wronskian is an upper diagonal. The determinant of an upper diagonal matrix is the product of the diagonal. We see that there can be no zero element on the diagonal. Hence the determinant is never zero. Therefore 1,x,x^{2},\cdots ,x^{n} are L.I.
Problem: Verify that y_{1}=x and y_{2}=x^{2} are L.I. solutions on the entire line of the equation x^{2}y^{\prime \prime }-2xy+2y=0 but that W\left ( x,x^{2}\right ) vanishes at x=0. Why does these observations not contradicts part (b) of theorem 3?
solution To verify that y_{1},y_{2} are solution of the ODE, we plugin each into the ODE and see if they satisfy it. For y_{1}, we obtain x^{2}y_{1}^{\prime \prime }-2xy_{1}+2y_{1}=0 But y_{1}^{\prime }=1,y_{1}^{\prime \prime }=0, therefore the above becomes\begin{align*} -2x+2x & =0\\ 0 & =0 \end{align*}
Verified. For y_{2}, where y_{2}^{\prime }=2x,y_{2}^{\prime \prime }=2, we obtain\begin{align*} x^{2}y_{2}^{\prime \prime }-2xy_{2}+2y_{2} & =0\\ 2x^{2}-4x^{2}+2x^{2} & =0\\ 0 & =0 \end{align*}
Hence verified. Now we need to show that y_{1},y_{2} are L.I. c_{1}y_{1}+c_{2}y_{2}=0 We now solve for c_{1},c_{2} c_{1}x+c_{2}x^{2}=0 Comparing coefficients on the LHS and RHS, we see that c_{1}=0,c_{2}=0. Hence this shows that y_{1},y_{2} are L.I. We now find the Wronskian W\left ( x\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} x & x^{2}\\ 1 & 2x \end{vmatrix} =2x^{2}-x^{2}=x^{2} Hence W\left ( x\right ) =0 at x=0. This does not contradicts part (b) of theorem 3, because when we write the ODE in the standard form y_{1}^{\prime \prime }-\frac{2}{x}y_{1}+\frac{2}{x^{2}}y_{1}=0 We see that p_{1}\left ( x\right ) =-\frac{2}{x},p_{2}\left ( x\right ) =\frac{2}{x^{2}}. These functions are not continuous at x=0 (there is singularity at x=0). But theorem 3 applies only to the interval where p_{i}\left ( x\right ) are continuous. Hence does not apply in this case. If W\left ( x\right ) was zero at location other than x=0, only then this will be a contradiction.
Problem: Find the general solution of the ODE y^{\prime \prime }+8y^{\prime }+25y=0
solution This is constant coefficient, linear, second order ODE. The characteristic equation is r^{2}+8r+25=0. The roots (using quadratic formula) are \begin{align*} r_{1} & =-4+3i\\ r_{2} & =-4-3i \end{align*}
Hence the general solution is\begin{align*} y & =c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}\\ & =c_{1}e^{\left ( -4+3i\right ) x}+c_{2}e^{\left ( -4-3i\right ) x}\\ & =e^{-4x}\left ( c_{1}\cos 3x+c_{2}\sin 3x\right ) \end{align*}
Problem: Find the general solution of the ODE y^{\left ( 4\right ) }+18y^{\prime \prime }+81y=0
solution This is constant coefficient, linear, second order ODE. The characteristic equation is r^{4}+18r^{2}+81=0. Let r^{2}=z, hence z^{2}+18z+81=0. This can be factored to \left ( z+9\right ) ^{2}=0. Hence the roots are -9 repeated. Therefore r^{2}=-9 or r=\pm 3i. Therefore, the 4 roots are \left \{ 3i,-3i,3i,-3i\right \} Hence the solution is y=c_{1}e^{3ix}+c_{2}e^{-3ix}+c_{3}xe^{3ix}+c_{4}xe^{-3ix} Or\begin{align*} y & =c_{1}\cos 3x+c_{2}\sin 3x+x\left ( c_{3}e^{3ix}+c_{4}e^{-3ix}\right ) \\ & =c_{1}\cos 3x+c_{2}\sin 3x+x\left ( c_{3}\cos 3x+c_{4}\sin 3x\right ) \end{align*}
Or y=\left ( c_{1}+xc_{3}\right ) \cos 3x+c_{2}\sin 3x\left ( c_{2}+xc_{4}\right )
Problem: Solve the initial value problem y^{\prime \prime }-6y^{\prime }+25y=0,y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =1
solution This is constant coefficient, linear, second order ODE. The characteristic equation is r^{2}-6r+25=0. Using quadratic formula, the roots are \begin{align*} r_{1} & =3+4i\\ r_{2} & =3-4i \end{align*}
Hence the general solution is y\left ( x\right ) =e^{3x}\left ( c_{1}\cos 4x+c_{2}\sin 4x\right ) Hence y^{\prime }\left ( x\right ) =3e^{3x}\left ( c_{1}\cos 4x+c_{2}\sin 4x\right ) +e^{3x}\left ( -c_{1}4\sin 4x+c_{2}4\cos 4x\right ) Applying first initial conditions gives 3=c_{1} Applying second initial conditions gives\begin{align*} 1 & =3\left ( 3\right ) +4c_{2}\\ c_{2} & =\frac{1-9}{4}=-2 \end{align*}
Hence the solution is y\left ( x\right ) =e^{3x}\left ( 3\cos 4x-2\sin 4x\right )
Problem: Solve the initial value problem y^{\left ( 3\right ) }+10y^{\prime \prime }+25y^{\prime }=0,y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =4,y^{\prime \prime }\left ( 0\right ) =5
solution This is constant coefficient, linear, second order ODE. The characteristic equation is r^{3}+10r^{2}+25r=0 or r\left ( r^{2}+10r+25\right ) =0, or r\left ( r+5\right ) ^{2}=0. Hence the roots are \left \{ 0,-5,-5\right \} . There are repeated root. Hence the solution is\begin{align*} y\left ( x\right ) & =c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}xe^{r_{2}x}\\ & =c_{1}+c_{2}e^{-5x}+c_{3}xe^{-5x} \end{align*}
Hence y^{\prime }=-5c_{2}e^{-5x}+c_{3}\left ( e^{-5x}-5xe^{-5x}\right ) And y^{\prime \prime }=25c_{2}e^{-5x}+c_{3}\left ( -5e^{-5x}-5\left ( e^{-5x}-5xe^{-5x}\right ) \right ) Applying first IC gives 3=c_{1}+c_{2} Applying second IC gives 4=-5c_{2}+c_{3} Applying third IC gives\begin{align*} 5 & =25c_{2}+c_{3}\left ( -5-5\right ) \\ & =25c_{2}-10c_{3} \end{align*}
We have three equations to solve for c_{1},c_{2},c_{3}.\begin{pmatrix} 1 & 1 & 0\\ 0 & -5 & 1\\ 0 & 25 & -10 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 3\\ 4\\ 5 \end{pmatrix} Therefore R_{3}=R_{3}+5R_{2} gives\begin{pmatrix} 1 & 1 & 0\\ 0 & -5 & 1\\ 0 & 0 & -5 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 3\\ 4\\ 25 \end{pmatrix} Therefore, from third row, -5c_{3}=25 or c_{3}=-5 From second row\begin{align*} -5c_{2}+c_{3} & =4\\ -5c_{2} & =4+5=9\\ c_{2} & =-\frac{9}{5} \end{align*}
From first row
\begin{align*} c_{1}+c_{2} & =3\\ c_{1} & =3+\frac{9}{5}\\ & =\frac{24}{5} \end{align*}
Hence the general solution is\begin{align*} y & =c_{1}+c_{2}e^{-5x}+c_{3}xe^{-5x}\\ & =\frac{24}{5}-\frac{9}{5}e^{-5x}-5xe^{-5x}\\ & =\frac{1}{5}\left ( 24-9e^{-5x}-25xe^{-5x}\right ) \end{align*}
Problem: One solution of the differential equation is given, find the second solution 6y^{\left ( 4\right ) }+5y^{\left ( 3\right ) }+25y^{\prime \prime }+20y^{\prime }+4y=0, and y_{1}=\cos 2x
solution The characteristic equation is 6r^{4}+5r^{3}+25r^{2}+20r+4=0\,. Since \cos 2x is solution, then this implies the roots for this solution must be \,r=\pm 2i, since this is what will give \cos 2x solution. Therefore, there must be factor \left ( r^{2}+4\right ) . Doing long division
\frac{6r^{4}+5r^{3}+25r^{2}+20r+4}{\left ( r^{2}+4\right ) }=6r^{2}+5r+1 Hence characteristic equation is \begin{align*} & \left ( r^{2}+4\right ) \left ( 6r^{2}+5r+1\right ) \\ & \left ( r^{2}+4\right ) \left ( 2r+1\right ) \left ( 3r+1\right ) \end{align*}
Hence the roots are r_{1}=2i,r_{2}=-2i,r_{3}=\frac{-1}{2},r_{4}=\frac{-1}{3}. Therefore the solution is\begin{align*} y & =c_{1}e^{2ix}+c_{2}e^{-2ix}+c_{3}e^{-\frac{1}{2}x}+c_{4}e^{\frac{-1}{3}x}\\ & =c_{1}\cos 2x+c_{2}\sin 2x+c_{3}e^{-\frac{1}{2}x}+c_{4}e^{\frac{-1}{3}x} \end{align*}
Problem: Solve y^{\left ( 3\right ) }-5y^{\prime \prime }+100y^{\prime }-500y=0 with y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =10,y^{\prime \prime }\left ( 0\right ) =250 given that y_{1}\left ( x\right ) =e^{5x} is one particular solution of the differential equation.
solution The characteristic equation is r^{3}-5r^{2}+100r-500=0. Since y_{1}\left ( x\right ) =e^{5x} is one solution, then \left ( r-5\right ) is one of the roots. Hence by long division \frac{r^{3}-5r^{2}+100r-500}{r-5}=r^{2}+100 Therefore the factoring of the characteristic equation is \left ( r-5\right ) \left ( r^{2}+100\right ) =0 Therefore the roots are r_{1}=5,r_{2}=10i,r_{3}=-10i and therefore the solution is y=c_{1}e^{5x}+c_{2}\cos 10x+c_{3}\sin 10x Hence\begin{align*} y^{\prime } & =5c_{1}e^{5x}-10c_{2}\sin 10x+10c_{3}\cos 10x\\ y^{\prime \prime } & =25c_{1}e^{5x}-100c_{2}\cos 10x-100c_{3}\sin 10x \end{align*}
Applying first IC gives 0=c_{1}+c_{2} Second IC gives 10=5c_{1}+10c_{3} Third IC gives 250=25c_{1}-100c_{2} We have three equations to solve for c_{1},c_{2},c_{3}.\begin{pmatrix} 1 & 1 & 0\\ 5 & 0 & 10\\ 25 & -100 & 0 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 10\\ 250 \end{pmatrix} R_{2}=R_{2}-5R_{1} and R_{3}=R_{3}-25R_{1} gives\begin{pmatrix} 1 & 1 & 0\\ 0 & -5 & 10\\ 0 & -125 & 0 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 10\\ 250 \end{pmatrix} R_{3}=R_{3}-25R_{2} gives\begin{pmatrix} 1 & 1 & 0\\ 0 & -5 & 10\\ 0 & 0 & -250 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 10\\ 0 \end{pmatrix} Hence from last row c_{3}=0. From second row -5c_{2}=10 or c_{2}=-2 and from first row c_{1}+c_{2}=0 or c_{1}=2. Hence the solution is\begin{align*} y & =c_{1}e^{5x}+c_{2}\cos 10x+c_{3}\sin 10x\\ & =2e^{5x}-2\cos 10x \end{align*}
Problem: Find linear homogeneous constant coefficient equation with the given general solution y\left ( x\right ) =Ae^{2x}+B\cos 2x+C\sin 2x
solution From the solution, we see that the roots are r_{1}=2,r_{2}=2i,r_{3}=-2i. Hence the characteristic equation is\begin{align*} \left ( r-2\right ) \left ( r^{2}+4\right ) & =0\\ r^{3}-2r^{2}+4r-8 & =0 \end{align*}
Therefore the original ODE is y^{\prime \prime \prime }\left ( x\right ) -2y^{\prime \prime }+4y^{\prime }-8y=0
Problem: Solve y^{\left ( 4\right ) }-y^{\left ( 3\right ) }-y^{\prime \prime }-y^{\prime }-2y=0 with y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0,y^{\prime \prime }\left ( 0\right ) =0,y^{\left ( 3\right ) }=30
solution The characteristic equation is r^{4}-r^{3}-r^{2}-r-2=0 By inspection, we see that r=-1 is a root. Hence by long division, we have \frac{r^{4}-r^{3}-r^{2}-r-2}{\left ( r+1\right ) }=r^{3}-2r^{2}+r-2 Therefore characteristic equation is \left ( r+1\right ) \left ( r^{3}-2r^{2}+r-2\right ) =0 By inspection, one of the roots of r^{3}-2r^{2}+r-2=0 is 2, hence by long division \frac{r^{3}-2r^{2}+r-2}{r-2}=r^{2}+1 Therefore characteristic equation becomes \left ( r+1\right ) \left ( r-2\right ) \left ( r^{2}+1\right ) =0 Hence roots are r_{1}=-1,r_{2}=2,r_{3}=-i,r_{4}=i and therefore the solution is y=c_{1}e^{-x}+c_{2}e^{2x}+c_{3}\cos x+c_{4}\sin x Initial conditions are now applied to find the constants.\begin{align*} y^{\prime } & =-c_{1}e^{-x}+2c_{2}e^{2x}-c_{3}\sin x+c_{4}\cos x\\ y^{\prime \prime } & =c_{1}e^{-x}+4c_{2}e^{2x}-c_{3}\cos x-c_{4}\sin x\\ y^{\prime \prime \prime } & =-c_{1}e^{-x}+8c_{2}e^{2x}+c_{3}\sin x-c_{4}\cos x \end{align*}
From y\left ( 0\right ) =0 we obtain 0=c_{1}+c_{2}+c_{3} From y^{\prime }\left ( 0\right ) =0 0=-c_{1}+2c_{2}+c_{4} From y^{\prime \prime }\left ( 0\right ) =0 0=c_{1}+4c_{2}-c_{3} And from y^{\prime \prime \prime }\left ( 0\right ) =30 30=-c_{1}+8c_{2}-c_{4} The 4 equations are solved for c_{i}\begin{pmatrix} 1 & 1 & 1 & 0\\ -1 & 2 & 0 & 1\\ 1 & 4 & -1 & 0\\ -1 & 8 & 0 & -1 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 30 \end{pmatrix} R_{2}=R_{2}+R_{1},R_{3}=R_{3}-R_{1},R_{4}=R_{4}+R_{1} gives\begin{pmatrix} 1 & 1 & 1 & 0\\ 0 & 3 & 1 & 1\\ 0 & 3 & -2 & 0\\ 0 & 9 & 1 & -1 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 30 \end{pmatrix} R_{3}=R_{3}-R_{2},R_{4}=R_{4}-3R_{2} gives\begin{pmatrix} 1 & 1 & 1 & 0\\ 0 & 3 & 1 & 1\\ 0 & 0 & -3 & -1\\ 0 & 0 & -2 & -4 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 30 \end{pmatrix} R_{4}=R_{4}-\frac{2}{3}R_{3} gives\begin{pmatrix} 1 & 1 & 1 & 0\\ 0 & 3 & 1 & 1\\ 0 & 0 & -3 & -1\\ 0 & 0 & 0 & -\frac{10}{3}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 30 \end{pmatrix} Hence from last row -\frac{10}{3}c_{4}=30,then c_{4}=-9 From 3rd row\begin{align*} -3c_{3}-c_{4} & =0\\ c_{3} & =3 \end{align*}
From second row\begin{align*} 3c_{2}+c_{3}+c_{4} & =0\\ 3c_{2}+3-9 & =0\\ c_{2} & =2 \end{align*}
From first row\begin{align*} c_{1}+c_{2}+c_{3} & =0\\ c_{1}+2+3 & =0\\ c_{1} & =-5 \end{align*}
Hence solution is\begin{align*} y & =c_{1}e^{-x}+c_{2}e^{2x}+c_{3}\cos x+c_{4}\sin x\\ & =-5e^{-x}+2e^{2x}+3\cos x-9\sin x \end{align*}