Solution
The ode to solve is \begin {align} x^{\prime \prime }\relax (t) +2x^{\prime }\relax (t) +x\left ( t\right ) & =0\tag {1}\\ x\relax (0) & =1\nonumber \\ x^{\prime }\relax (0) & =0\nonumber \end {align}
This is a constant coefficient ODE. Assuming the solution has the form \(x=Ae^{\lambda t}\) and substituting this back in (1) gives the characteristic equation (the constant \(A\) drops out)\begin {align*} \lambda ^{2}e^{\lambda t}+2\lambda e^{\lambda t}+e^{\lambda t} & =0\\ \left (\lambda ^{2}+42\lambda +1\right ) e^{\lambda t} & =0 \end {align*}
Since \(e^{\lambda t}\neq 0\), the above gives\begin {align*} \lambda ^{2}+2\lambda +1 & =0\\ \left (\lambda +1\right ) ^{2} & =0 \end {align*}
Therefore \(\lambda =-1\). (double root). Since the root is double, then the basis solutions are \(x_{1}\relax (t) =e^{\lambda t},x_{2}\left ( t\right ) =te^{\lambda t}\) and the general solution is a linear combination of these basis solutions. Therefore the general solution is\begin {equation} x\relax (t) =Ae^{-t}+Bte^{-t} \tag {2} \end {equation} The constants \(A,B\) are found from initial conditions. At \(t=0\) and using \(x\relax (0) =1\) gives\begin {equation} 1=A \tag {3} \end {equation} Solution (2) becomes\begin {equation} x\relax (t) =e^{-t}+Bte^{-t} \tag {4} \end {equation} Taking derivative of (4) gives\[ x^{\prime }\relax (t) =-e^{-t}+Be^{t}-Bte^{-t}\] Using \(x^{\prime }\relax (0) =0\) on the above gives\begin {align} 0 & =-1+B\nonumber \\ B & =1 \tag {5} \end {align}
Substituting (3,5) in (4) gives the final solution\begin {align*} x\relax (t) & =e^{-t}+te^{-t}\\ & =\left (1+t\right ) e^{-t} \end {align*}
The ode to solve is \[ x^{\prime \prime \prime \prime }\relax (t) +x\relax (t) =0 \] As was done in the above part, substituting \(x=Ae^{\lambda t}\) in the above and simplifying gives the characteristic equation\[ \lambda ^{4}+1=0 \] Hence the roots are \(\lambda ^{4}=-1\) or \(\lambda ^{4}=e^{-i\frac {\pi }{2}}\). There are 4 roots that divide the unit circle equally, each is \(90\) degrees phase shifted (anti clockwise)from the other, starting from first root at phase \(-\frac {\pi }{2}=-45\) degrees. Hence the roots are\begin {align*} \lambda _{1} & =\cos \left (-45\right ) +i\sin \left (-45\right ) \\ \lambda _{2} & =\cos \left (45\right ) +i\sin \left (45\right ) \\ \lambda _{3} & =\cos \left (135\right ) +i\sin \left (135\right ) \\ \lambda _{4} & =\cos \left (225\right ) +i\sin \left (225\right ) \end {align*}
or\begin {align*} \lambda _{1} & =\frac {\sqrt {2}}{2}-i\frac {\sqrt {2}}{2}\\ \lambda _{2} & =\frac {\sqrt {2}}{2}+i\frac {\sqrt {2}}{2}\\ \lambda _{3} & =-\frac {\sqrt {2}}{2}+i\frac {\sqrt {2}}{2}\\ \lambda _{4} & =-\frac {\sqrt {2}}{2}-i\frac {\sqrt {2}}{2} \end {align*}
Therefore the basis solutions are \begin {align*} x_{1}\relax (t) & =e^{\left (\frac {\sqrt {2}}{2}-i\frac {\sqrt {2}}{2}\right ) t}\\ x_{2}\relax (t) & =e^{\left (\frac {\sqrt {2}}{2}+i\frac {\sqrt {2}}{2}\right ) t}\\ x_{3}\relax (t) & =e^{\left (-\frac {\sqrt {2}}{2}+i\frac {\sqrt {2}}{2}\right ) t}\\ x_{4}\relax (t) & =e^{\left (-\frac {\sqrt {2}}{2}-i\frac {\sqrt {2}}{2}\right ) t} \end {align*}
The general solution is linear combination of the above basis solutions, which becomes\begin {align*} x\relax (t) & =c_{1}e^{\left (\frac {\sqrt {2}}{2}-i\frac {\sqrt {2}}{2}\right ) t}+c_{2}e^{\left (\frac {\sqrt {2}}{2}+i\frac {\sqrt {2}}{2}\right ) t}+c_{3}e^{\left (-\frac {\sqrt {2}}{2}+i\frac {\sqrt {2}}{2}\right ) t}+c_{4}e^{\left (-\frac {\sqrt {2}}{2}-i\frac {\sqrt {2}}{2}\right ) t}\\ & =c_{1}e^{\frac {\sqrt {2}}{2}t}e^{-i\frac {\sqrt {2}}{2}t}+c_{2}e^{\frac {\sqrt {2}}{2}t}e^{i\frac {\sqrt {2}}{2}t}+c_{3}e^{-\frac {\sqrt {2}}{2}t}e^{i\frac {\sqrt {2}}{2}t}+c_{4}e^{-\frac {\sqrt {2}}{2}t}e^{-i\frac {\sqrt {2}}{2}t}\\ & =e^{\frac {\sqrt {2}}{2}t}\left (c_{1}e^{-i\frac {\sqrt {2}}{2}t}+c_{2}e^{i\frac {\sqrt {2}}{2}t}\right ) +e^{-\frac {\sqrt {2}}{2}t}\left ( c_{3}e^{i\frac {\sqrt {2}}{2}t}+c_{4}e^{-i\frac {\sqrt {2}}{2}t}\right ) \end {align*}
Using Euler relation, the above can be rewritten as\[ x\relax (t) =e^{\frac {\sqrt {2}}{2}t}\left (c_{1}\sin \left ( \frac {\sqrt {2}}{2}t\right ) +c_{2}\cos \left (\frac {\sqrt {2}}{2}t\right ) \right ) +e^{-\frac {\sqrt {2}}{2}t}\left (c_{3}\sin \left (\frac {\sqrt {2}}{2}t\right ) +c_{4}\cos \left (\frac {\sqrt {2}}{2}t\right ) \right ) \]
The ode to solve is \[ x^{\prime \prime \prime }\relax (t) -3x^{\prime \prime }\relax (t) -9x^{\prime }\relax (t) -5x\relax (t) =0 \] As was done in the above part, substituting \(x=Ae^{\lambda t}\) in the above and simplifying gives the characteristic equation\[ \lambda ^{3}-3\lambda ^{2}-9\lambda -5=0 \] Since one root is \(5\), then the above can be written as\[ \left (\lambda -5\right ) \left (\Delta \right ) =0 \] Where \[ \Delta =\frac {\lambda ^{3}-3\lambda ^{2}-9\lambda -5}{\lambda -5}\] Using long division gives\[ \Delta =\left (\lambda +1\right ) ^{2}\] Therefore the roots of the characteristic equation are\begin {align*} \lambda _{1} & =5\\ \lambda _{2} & =-1\\ \lambda _{3} & =-1 \end {align*}
roots \(\lambda _{2},\lambda _{3}\) are the same. \(\lambda =-1\) is a double root. Therefore the basis solutions are\begin {align*} x_{1}\relax (t) & =e^{5t}\\ x_{2}\relax (t) & =e^{t}\\ x_{3}\relax (t) & =te^{t} \end {align*}
Where \(t\) multiplies the last basis \(x_{3}\relax (t) \) due to the double root. The general solution is linear combination of the above basis solutions, which gives\begin {align*} x\relax (t) & =c_{1}x_{1}\relax (t) +c_{2}x_{2}\left ( t\right ) +c_{3}x_{3}\relax (t) \\ & =c_{1}e^{5t}+c_{2}e^{t}+c_{3}te^{t} \end {align*}
The ode to solve is \[ \left (D+1\right ) ^{2}\left (D^{4}-256\right ) x\relax (t) =0 \] This has the characteristic equation equation \(\left (\lambda +1\right ) ^{2}\left (\lambda ^{4}-256\right ) =0\). The roots of \(\left (\lambda ^{4}-256\right ) \) are given by \(\lambda ^{4}=256\). Let \(\lambda ^{2}=\omega \). Therefore \(\omega ^{2}=256\) which gives \(\omega =\pm 16\).
When \(\omega =16\), then \(\lambda ^{2}=16\) which gives \(\lambda =\pm 4\) and when \(\omega =-16\), then \(\lambda ^{2}=-16\) which gives \(\lambda =\pm 4i.\)
The other part \(\left (\lambda +1\right ) ^{2}=0\) gives \(\lambda =-1\), double root. Therefore the roots of the characteristic equation are\begin {align*} \lambda _{1} & =4\\ \lambda _{2} & =-4\\ \lambda _{3} & =4i\\ \lambda _{4} & =-4i\\ \lambda _{5} & =-1\\ \lambda _{6} & =-1 \end {align*}
Root \(\lambda =-1\) is a double root. Therefore the basis solutions as\begin {align*} x_{1}\relax (t) & =e^{4t}\\ x_{2}\relax (t) & =e^{-4t}\\ x_{3}\relax (t) & =e^{4it}\\ x_{4}\relax (t) & =e^{-4it}\\ x_{5}\relax (t) & =e^{-t}\\ x_{6}\relax (t) & =te^{-t} \end {align*}
Where \(t\) was multiplied by \(e^{-t}\) in \(x_{6}\relax (t) \) since the root is double. The solution is linear combination of the above basis solutions, which gives\begin {align*} x\relax (t) & =c_{1}x_{1}\relax (t) +c_{2}x_{2}\left ( t\right ) +c_{3}x_{3}\relax (t) +c_{4}x_{4}\relax (t) +c_{5}x_{5}\relax (t) +c_{6}x_{6}\relax (t) \\ & =c_{1}e^{4t}+c_{2}e^{-4t}+c_{3}e^{4it}+c_{4}e^{-4it}+c_{5}e^{-t}+c_{6}te^{-t}\\ & =e^{-t}\left (c_{5}+tc_{6}\right ) +c_{1}e^{4t}+c_{2}e^{-4t}+c_{3}\sin \left (4t\right ) +c_{4}\cos \left (4t\right ) \end {align*}
Where Euler relation was used in the last step above to rewrite \(c_{3}e^{4it}+c_{4}e^{-4it}\).
Solution
The ode to solve is \begin {equation} y^{\prime \prime }-y^{\prime }-2y=e^{2x} \tag {1} \end {equation} This is second order constant coefficients inhomogeneous ODE. The general solution is \begin {equation} y\relax (x) =y_{h}\relax (x) +y_{p}\relax (x) \tag {2} \end {equation} Where \(y_{h}\relax (x) \) is the solution to \(y^{\prime \prime }-y^{\prime }-2y=0\) and \(y_{p}\relax (x) \) is any particular solution to \(y^{\prime \prime }-y^{\prime }-2y=e^{2x}\). The homogenous solution is found using the characteristic polynomial method as was done in the above problems. Substituting \(y=Ae^{\lambda x}\) in \(y^{\prime \prime }-y^{\prime }-2y=0\) and simplifying gives\begin {align*} \lambda ^{2}-\lambda -2 & =0\\ \left (\lambda +1\right ) \left (\lambda -2\right ) & =0 \end {align*}
The roots are \(\lambda _{1}=-1,\lambda _{2}=2\). Therefore the basis solutions are \begin {align} y_{1}\relax (x) & =e^{-x}\tag {3}\\ y_{1}\relax (x) & =e^{2x}\nonumber \end {align}
Hence \(y_{h}\relax (x) \) is linear combination of the above, which gives\[ y_{h}\relax (x) =c_{1}e^{-x}+c_{2}e^{2x}\] The particular solution is now found. Assuming \(y_{p}=Ae^{2x}\). But \(e^{2x}\) is a basis solution of the homogeneous ode. Therefore \(y_{p}\) is multiplied by \(x\) giving \[ y_{p}=Axe^{2x}\] Substituting this back in (1) and solving for \(A\) gives\begin {align*} y_{p}^{\prime } & =Ae^{2x}+2Axe^{2x}\\ y_{p}^{\prime \prime } & =2Ae^{2x}+2Ae^{2x}+4Axe^{2x}\\ & =4Ae^{2x}+4Axe^{2x} \end {align*}
Eq (1) becomes\begin {align*} \left (4Ae^{2x}+4Axe^{2x}\right ) -\left (Ae^{2x}+2Axe^{2x}\right ) -2\left (Axe^{2x}\right ) & =e^{2x}\\ 4Ae^{2x}+4Axe^{2x}-Ae^{2x}-2Axe^{2x}-2Axe^{2x} & =e^{2x}\\ 4A+4Ax-A-2Ax-2Ax & =1\\ 3A & =1\\ A & =\frac {1}{3} \end {align*}
Hence the particular solution is\[ y_{p}\relax (x) =\frac {1}{3}xe^{2x}\] Therefore from (2) the general solution is\begin {equation} y\relax (x) =c_{1}e^{-x}+c_{2}e^{2x}+\frac {1}{3}xe^{2x} \tag {4} \end {equation} \(c_{1},c_{2}\) are now found from initial conditions. At \(x=0\), (4) becomes\begin {equation} 1=c_{1}+c_{2} \tag {5} \end {equation} Taking derivative of (4) gives\[ y^{\prime }\relax (x) =-c_{1}e^{-x}+2c_{2}e^{2x}+\frac {1}{3}e^{2x}+\frac {2}{3}xe^{2x}\] At \(x=0\) the above gives\begin {equation} 0=-c_{1}+2c_{2}+\frac {1}{3} \tag {6} \end {equation} Eq (5,6) are now solved for \(c_{1},c_{2}\). From (5) \[ c_{1}=1-c_{2}\] Substituting this back in (6) gives\begin {align*} 0 & =-\left (1-c_{2}\right ) +2c_{2}+\frac {1}{3}\\ c_{2} & =\frac {2}{9} \end {align*}
Therefore \(c_{1}=1-\frac {2}{9}=\frac {7}{9}\). The final solution (4) becomes\[ y\relax (x) =\frac {7}{9}e^{-x}+\frac {2}{9}e^{2x}+\frac {1}{3}xe^{2x}\]
The ode to solve is \begin {equation} y^{\prime \prime }-2y^{\prime }+y=2\cos x \tag {1} \end {equation} This is second order constant coefficients inhomogeneous ODE. Hence the general solution is \begin {equation} y\relax (x) =y_{h}\relax (x) +y_{p}\relax (x) \tag {2} \end {equation} Where \(y_{h}\relax (x) \) is the solution to \(y^{\prime \prime }-2y^{\prime }+y=0\) and \(y_{p}\relax (x) \) is any particular solution to \(y^{\prime \prime }-2y^{\prime }+y=2\cos x\). The homogenous is found using the characteristic polynomial method. Substituting \(y=Ae^{\lambda x}\) in \(y^{\prime \prime }-2y^{\prime }+y=0\) and simplifying gives\begin {align*} \lambda ^{2}-2\lambda +1 & =0\\ \left (\lambda -1\right ) \left (\lambda -1\right ) & =0 \end {align*}
roots are \(\lambda _{1}=1,\lambda _{2}=1\). (double root). The basis solutions are therefore\begin {align} y_{1}\relax (x) & =e^{x}\tag {3}\\ y_{1}\relax (x) & =xe^{x}\nonumber \end {align}
\(y_{h}\relax (x) \) is linear combination of the the above which gives\[ y_{h}\relax (x) =c_{1}e^{x}+c_{2}xe^{x}\] The particular solution is now found. Assuming \(y_{p}=A\cos x\). Taking all derivatives of this solution gives the set \(\left \{ \cos x,\sin x\right \} \). Therefore \[ y_{p}=A\cos x+B\sin x \] Substituting this back in (1) to solve for \(A,B\) gives\begin {align*} y_{p}^{\prime } & =-A\sin x+B\cos x\\ y_{p}^{\prime \prime } & =-A\cos x-B\sin x \end {align*}
Hence (1) becomes\begin {align*} y_{p}^{\prime \prime }-2y_{p}^{\prime }+y_{p} & =2\cos x\\ \left (-A\cos x-B\sin x\right ) -2\left (-A\sin x+B\cos x\right ) +\left ( A\cos x+B\sin x\right ) & =2\cos x\\ -A\cos x-B\sin x+2A\sin x-2B\cos x+A\cos x+B\sin x & =2\cos x\\ \cos x\left (-A-2B+A\right ) +\sin x\left (-B+2A+B\right ) & =2\cos x\\ -2B\cos x+2A\sin x & =2\cos x \end {align*}
Hence \(A=0\) and \(B=-1\). Therefore the particular solution is\[ y_{p}\relax (x) =-\sin x \] Eq (2) becomes\begin {equation} y\relax (x) =c_{1}e^{x}+c_{2}xe^{x}-\sin x \tag {4} \end {equation} \(c_{1},c_{2}\) are now found from initial conditions. At \(x=0\), (4) becomes\begin {equation} 1=c_{1} \tag {5} \end {equation} The solution (4) becomes\begin {equation} y\relax (x) =e^{x}+c_{2}xe^{x}-\sin x \tag {6} \end {equation} Taking derivative of (6) gives\[ y^{\prime }\relax (x) =e^{x}+c_{2}e^{x}+c_{2}xe^{x}-\cos x \] At \(x=0\) the above gives\begin {equation} 0=1+c_{2}-1 \tag {6} \end {equation} Therefore \(c_{2}=0\) and now Eq (6) gives the final solution as\[ y\relax (x) =e^{x}-\sin x \]
The ode to solve is \begin {equation} y^{\prime \prime }+16y=16\cos 4x \tag {1} \end {equation} This is second order constant coefficients inhomogeneous ODE. Hence the general solution is \begin {equation} y\relax (x) =y_{h}\relax (x) +y_{p}\relax (x) \tag {2} \end {equation} Where \(y_{h}\relax (x) \) is the solution to \(y^{\prime \prime }+16y=0\) and \(y_{p}\relax (x) \) is any particular solution to \(y^{\prime \prime }+16y=16\cos 4x\). The homogenous is found using the characteristic polynomial method. Substituting \(y=Ae^{\lambda x}\) in \(y^{\prime \prime }+16y=0\) and simplifying gives\begin {align*} \lambda ^{2}+16 & =0\\ \lambda & =\pm 4i \end {align*}
The roots are \(\lambda _{1}=4i,\lambda _{2}=-4i\). The basis solutions are therefore\begin {align} y_{1}\relax (x) & =e^{i4x}\tag {3}\\ y_{2}\relax (x) & =e^{-i4x}\nonumber \end {align}
Therefore \(y_{h}\relax (x) \) is linear combination of the the above. \[ y_{h}\relax (x) =c_{1}e^{i4x}+c_{2}e^{-i4x}\] Which can be written, using Euler formula as\[ y_{h}\relax (x) =c_{1}\cos 4x+c_{2}\sin 4x \] The particular solution is now found. Assuming \(y_{p}=A\cos 4x\). Taking all derivatives of this, the basis for \(y_{p}\) becomes \(\left \{ \cos 4x,\sin 4x\right \} \). But \(\cos 4x\) is a basis of \(y_{h}\). Therefore this set is multiplied by \(x\). The whole set is multiplied by \(x\) and not just \(\cos 4x\) because the set was generated by taking derivative of \(\cos 4x\).
The basis set for \(y_{p}\) now becomes \(\left \{ x\cos 4x,x\sin 4x\right \} \). Hence \(y_{p}\) is linear combination of these basis, giving trial \(y_{p}\) as\begin {equation} y_{p}=Ax\cos 4x+Bx\sin 4x \tag {4} \end {equation} Therefore\begin {align*} y_{p}^{\prime } & =\left (A\cos 4x-4Ax\sin 4x\right ) +\left (B\sin 4x+4Bx\cos 4x\right ) \\ y_{p}^{\prime \prime } & =\left (-4A\sin 4x-4A\sin 4x-16Ax\cos 4x\right ) +\left (4B\cos 4x+4B\cos 4x-16Bx\sin 4x\right ) \\ & =-8A\sin 4x-16Ax\cos 4x+8B\cos 4x-16Bx\sin 4x \end {align*}
Substituting the above back in (1) gives\begin {align*} \left (-8A\sin 4x-16Ax\cos 4x+8B\cos 4x-16Bx\sin 4x\right ) +16\left ( Ax\cos 4x+Bx\sin 4x\right ) & =16\cos 4x\\ \sin 4x\left (-8A-16Bx+16Bx\right ) +\cos 4x\left (-16Ax+8B+16Ax\right ) & =16\cos 4x \end {align*}
Hence\begin {align*} -16Ax+8B+16Ax & =16\\ -8A-16Bx+16Bx & =0 \end {align*}
Or\begin {align*} 8B & =16\\ -8A & =0 \end {align*}
First equation gives \(B=2\). Second equation gives \(A=0\). Therefore the particular solution (4) becomes\[ y_{p}=2x\sin 4x \] From (2), the general solution becomes\begin {align} y\relax (x) & =y_{h}\relax (x) +y_{p}\relax (x) \nonumber \\ & =c_{1}\cos 4x+c_{2}\sin 4x+2x\sin 4x \tag {5} \end {align}
\(c_{1},c_{2}\) are now found from initial conditions. At \(x=0\), (5) becomes\[ 1=c_{1}\] Hence the solution (5) becomes\begin {equation} y\relax (x) =\cos 4x+c_{2}\sin 4x+2x\sin 4x \tag {6} \end {equation} Taking derivative of the above\[ y^{\prime }\relax (x) =-4\sin 4x+4c_{2}\cos 4x+2\sin 4x+8x\cos 4x \] At \(t=0\) the above gives\[ 0=4c_{2}\] Hence \(c_{2}=0\) and the final solution (6) becomes\[ y\relax (x) =\cos 4x+2x\sin 4x \]
The ode to solve is \begin {equation} y^{\prime \prime }-y=\cosh x \tag {1} \end {equation} This is second order constant coefficients inhomogeneous ODE. Hence the general solution is \begin {equation} y\relax (x) =y_{h}\relax (x) +y_{p}\relax (x) \tag {2} \end {equation} Where \(y_{h}\relax (x) \) is the solution to \(y^{\prime \prime }+y=0\) and \(y_{p}\relax (x) \) is any particular solution to \(y^{\prime \prime }+y=\cosh x\). The homogenous is found using the characteristic polynomial method. Substituting \(y=Ae^{\lambda x}\) in \(y^{\prime \prime }+y=0\) and simplifying gives\begin {align*} \lambda ^{2}-1 & =0\\ \lambda & =\pm 1 \end {align*}
roots are \(\lambda _{1}=1,\lambda _{2}=-1\). The basis solutions are therefore\begin {align} y_{1}\relax (x) & =e^{x}\tag {3}\\ y_{2}\relax (x) & =e^{-x}\nonumber \end {align}
Therefore \(y_{h}\relax (x) \) is linear combination of the the above. \[ y_{h}\relax (x) =c_{1}e^{x}+c_{2}e^{-x}\] Which can be written, using Euler formula as\[ y_{h}\relax (x) =c_{1}\cosh x+c_{2}\sinh x \] The particular solution is now found. Assuming \(y_{p}=A\cosh x\). Taking all derivatives of this, the basis for \(y_{p}\) becomes \(\left \{ \cosh x,\sinh x\right \} \). But \(\cosh x\) is basis of \(y_{h}\). Therefore this set is multiplied by \(x\). The whole set is multiplied by \(x\) and not just \(\cosh x\) because the set was generated by taking derivative of \(\cosh x\).
The basis set for \(y_{p}\) becomes \(\left \{ x\cosh x,x\sinh x\right \} \). Hence \(y_{p}\) is linear combination of these basis, giving trial \(y_{p}\) as\begin {equation} y_{p}=Ax\cosh x+Bx\sinh x \tag {4} \end {equation} Therefore\begin {align*} y_{p}^{\prime } & =A\cosh x+Ax\sinh x+B\sinh x+Bx\cosh x\\ y_{p}^{\prime \prime } & =A\sinh x+A\sinh x+Ax\cosh x+B\cosh x+B\cosh x+Bx\sinh x\\ & =2A\sinh x+Ax\cosh x+2B\cosh x+Bx\sinh x \end {align*}
Substituting the above back in (1) gives\begin {align*} \left (2A\sinh x+Ax\cosh x+2B\cosh x+Bx\sinh x\right ) -\left (Ax\cosh x+Bx\sinh x\right ) & =\cosh x\\ \sinh x\left (2A+Bx-Bx\right ) +\cosh x\left (Ax+2B-Ax\right ) & =\cosh x \end {align*}
Hence\begin {align*} 2B & =1\\ 2A & =0 \end {align*}
Therefore \(B=\frac {1}{2},A=0\) and (4) becomes\[ y_{p}=\frac {1}{2}x\sinh x \] From (2), the general solution becomes\begin {align} y\relax (x) & =y_{h}\relax (x) +y_{p}\relax (x) \nonumber \\ & =c_{1}\cosh x+c_{2}\sinh x+\frac {1}{2}x\sinh x \tag {5} \end {align}
\(c_{1},c_{2}\) are now found from initial conditions. At \(x=0\), (5) becomes\[ 1=c_{1}\] Hence the solution (5) becomes\begin {equation} y\relax (x) =\cosh x+c_{2}\sinh x+\frac {1}{2}x\sinh x \tag {6} \end {equation} Taking derivative of the above\[ y^{\prime }\relax (x) =\sinh x+c_{2}\cosh x+\frac {1}{2}\sinh x+\frac {1}{2}x\cosh x \] At \(t=0\) the above gives\[ 0=c_{2}\cosh x \] Hence \(c_{2}=0\) and the final solution (6) becomes\[ y\relax (x) =\cosh x+\frac {1}{2}x\sinh x \]
Solve \(x^{2}y^{\prime }+2xy=\sinh x\) with \(y\relax (1) =2\)
Solution
Dividing by \(x\neq 0\)\[ y^{\prime }+2\frac {y}{x}=\frac {\sinh x}{x^{2}}\] The integrating factor is \(I=e^{\int \frac {2}{x}dx}=e^{2\ln x}=x^{2}\). Multiplying both sides by this integration factor makes the left side a complete differential\begin {align*} \frac {d}{dx}\left (yx^{2}\right ) & =x^{2}\frac {\sinh x}{x^{2}}\\ \frac {d}{dx}\left (yx^{2}\right ) & =\sinh x \end {align*}
Integrating gives\begin {align} yx^{2} & =\int \sinh xdx+C\nonumber \\ yx^{2} & =\cosh x+C\nonumber \\ y & =\frac {\cosh x}{x^{2}}+\frac {C}{x^{2}} \tag {1} \end {align}
At \(x=1\) the above becomes\begin {align*} 2 & =\cosh 1+C\\ C & =2-\cosh 1 \end {align*}
Hence the solution (1) becomes\begin {align*} y\relax (x) & =\frac {\cosh x}{x^{2}}+\frac {2-\cosh 1}{x^{2}}\\ & =\frac {1}{x^{2}}\left (\cosh x+2-\cosh 1\right ) \end {align*}
Where \(x\neq 0\)
Solve \[ \left (1+x^{2}\right ) y^{\prime }=1+xy \] Solution
\begin {align*} y^{\prime } & =\frac {1+xy}{1+x^{2}}\\ & =\frac {1}{1+x^{2}}+\frac {xy}{1+x^{2}} \end {align*}
Therefore\begin {equation} y^{\prime }-y\frac {x}{1+x^{2}}=\frac {1}{1+x^{2}} \tag {1} \end {equation} This is linear in \(y\) first order ODE. It has the form \(y^{\prime }+p\left ( x\right ) y=q\relax (x) \). The integration factor is \begin {align*} I & =e^{\int p\relax (x) dx}\\ & =e^{-\int \frac {x}{1+x^{2}}dx} \end {align*}
But \(\int \frac {x}{1+x^{2}}dx=\frac {1}{2}\ln \left (1+x^{2}\right ) \). Therefore\begin {align*} I & =e^{-\frac {1}{2}\ln \left (1+x^{2}\right ) }\\ & =e^{\ln \left (1+x^{2}\right ) ^{-\frac {1}{2}}}\\ & =\left (1+x^{2}\right ) ^{-\frac {1}{2}}\\ & =\frac {1}{\sqrt {1+x^{2}}} \end {align*}
Multiplying both sides of (1) by this integrating factor makes the left side a complete differential\begin {align*} \frac {d}{dx}\left (y\frac {1}{\sqrt {1+x^{2}}}\right ) & =\frac {1}{\sqrt {1+x^{2}}}\frac {1}{1+x^{2}}\\ \frac {d}{dx}\left (y\frac {1}{\sqrt {1+x^{2}}}\right ) & =\frac {1}{\left ( 1+x^{2}\right ) ^{\frac {3}{2}}}\\ & =\left (1+x^{2}\right ) ^{-\frac {3}{2}} \end {align*}
Integrating gives\begin {equation} y\frac {1}{\sqrt {1+x^{2}}}=\int \left (1+x^{2}\right ) ^{-\frac {3}{2}}dx+C \tag {2} \end {equation} To integrate \(\int \frac {1}{\left (1+x^{2}\right ) ^{\frac {3}{2}}}dx\), let \(x=\tan u\), then \(dx=\left (1+\tan ^{2}u\right ) du\). Hence\begin {align*} \int \frac {1}{\left (1+x^{2}\right ) ^{\frac {3}{2}}}dx & =\int \frac {1}{\left (1+\tan ^{2}u\right ) ^{\frac {3}{2}}}\left (1+\tan ^{2}u\right ) du\\ & =\int \frac {1}{\left (1+\tan ^{2}u\right ) ^{\frac {1}{2}}}du\\ & =\int \frac {1}{\left (1+\frac {\sin ^{2}u}{\cos ^{2}u}\right ) ^{\frac {1}{2}}}du\\ & =\int \frac {\cos u}{\left (\cos ^{2}u+\sin ^{2}u\right ) ^{\frac {1}{2}}}du\\ & =\int \cos u\ du\\ & =\sin u \end {align*}
But \(\sin u=\frac {\frac {\sin u}{\cos u}}{\sqrt {1+\frac {\sin ^{2}u}{\cos ^{2}u}}}=\frac {\tan u}{\sqrt {1+\tan ^{2}u}}=\frac {x}{\sqrt {1+x^{2}}}\). Hence \[ \int \frac {1}{\left (1+x^{2}\right ) ^{\frac {3}{2}}}dx=\frac {x}{\sqrt {1+x^{2}}}\] Therefore the final solution (2) becomes\begin {align} y\frac {1}{\sqrt {1+x^{2}}} & =\frac {x}{\sqrt {1+x^{2}}}+C\nonumber \\ y & =x+C\sqrt {1+x^{2}} \tag {3} \end {align}
Solve (a) \(y^{\prime }+xy=xy^{2}\) (b) \(3xy^{\prime }+y+x^{2}y^{4}=0\)
Solution
The ode has the form\[ y^{\prime }+p\relax (x) y=q\relax (x) y^{m}\] Where \(p\relax (x) =x,q\relax (x) =x\) and \(m=2\). Therefore this is Bernoulli ODE. The first step is to divide throughout by \(y^{m}=y^{2}\) which gives\begin {equation} \frac {y^{\prime }}{y^{2}}+p\relax (x) y^{-1}=q\relax (x) \tag {1} \end {equation} Setting \begin {equation} v\relax (x) =y^{-1} \tag {2} \end {equation} Taking derivatives of the above w.r.t. \(x\) gives \begin {equation} v^{\prime }\relax (x) =\frac {-1}{y^{2}}y^{\prime }\relax (x) \tag {3} \end {equation} Substituting (2,3) into (1) gives\[ -v^{\prime }\relax (x) +p\relax (x) v\relax (x) =q\left ( x\right ) \] But here \(p\relax (x) =x\) and \(q\relax (x) =x\). The above becomes\begin {align*} -v^{\prime }\relax (x) +xv\relax (x) & =x\\ v^{\prime }\relax (x) -xv\relax (x) & =-x \end {align*}
This is linear ODE in \(v\relax (x) \). The integrating factor is \(e^{\int -xdx}=e^{-\frac {x^{2}}{2}}\). Multiplying both sides of the above by this integrating factor makes the left side a complete differential\[ \frac {d}{dx}\left (ve^{-\frac {x^{2}}{2}}\right ) =-xe^{-\frac {x^{2}}{2}}\] Integrating gives\begin {equation} ve^{-\frac {x^{2}}{2}}=-\int xe^{-\frac {x^{2}}{2}}dx+C \tag {4} \end {equation} To integrate \(\int xe^{-\frac {x^{2}}{2}}dx\), let \(u=x^{2}\). Then \(du=2xdx\). Substituting gives\begin {align*} \int xe^{-\frac {x^{2}}{2}}dx & =\int xe^{-\frac {u}{2}}\frac {du}{2x}\\ & =\frac {1}{2}\int e^{-\frac {u}{2}}du\\ & =\frac {1}{2}\frac {e^{-\frac {u}{2}}}{-\frac {1}{2}}\\ & =-e^{-\frac {u}{2}} \end {align*}
But \(u=x^{2}\). Therefore\[ \int xe^{-\frac {x^{2}}{2}}dx=-e^{-\frac {x^{2}}{2}}\] Substituting the above in (4) gives\begin {align*} ve^{-\frac {x^{2}}{2}} & =e^{-\frac {x^{2}}{2}}+C\\ v & =1+e^{\frac {x^{2}}{2}}C \end {align*}
But \(v=y^{-1}\), therefore\begin {align*} y^{-1} & =1+e^{\frac {x^{2}}{2}}C\\ y\relax (x) & =\frac {1}{1+e^{\frac {x^{2}}{2}}C} \end {align*}
Where \(C\) is constant of integration.
The ode is\[ 3xy^{\prime }+y+x^{2}y^{4}=0 \] Dividing by \(3x\) for \(x\neq 0\) gives\begin {align*} y^{\prime }+\frac {y}{3x}+\frac {x}{3}y^{4} & =0\\ y^{\prime }+\frac {1}{3x}y & =-\frac {x}{3}y^{4} \end {align*}
Now this ODE has the Bernoulli form, \[ y^{\prime }+p\relax (x) y=q\relax (x) y^{m}\] Where \(p\relax (x) =\frac {1}{3x},q\relax (x) =-\frac {x}{3}\) and \(m=4\). Therefore this is Bernoulli ODE. The first step is to divide throughout by \(y^{m}=y^{4}\) which gives\begin {equation} \frac {y^{\prime }}{y^{4}}+p\relax (x) y^{-3}=q\relax (x) \tag {1} \end {equation} Setting \begin {equation} v\relax (x) =y^{-3} \tag {2} \end {equation} Taking derivatives of the above w.r.t. \(x\) gives \begin {equation} v^{\prime }\relax (x) =\frac {-3}{y^{4}}y^{\prime }\relax (x) \tag {3} \end {equation} Substituting (2,3) into (1) gives\[ -\frac {1}{3}v^{\prime }\relax (x) +p\relax (x) v\relax (x) =q\relax (x) \] But here \(p\relax (x) =\frac {1}{3x},q\relax (x) =-\frac {x}{3}\). The above becomes\begin {align*} -\frac {1}{3}v^{\prime }\relax (x) +\frac {1}{3x}v\relax (x) & =-\frac {x}{3}\\ v^{\prime }\relax (x) -\frac {1}{x}v\relax (x) & =x \end {align*}
This is linear in \(v\relax (x) \). The integrating factor is \(e^{-\int \frac {1}{x}dx}=e^{-\ln x}=\frac {1}{x}\). Multiplying both sides of the above by this integrating factor make the left side a complete differential\[ \frac {d}{dx}\left (v\frac {1}{x}\right ) =1 \] Integrating gives\begin {align} v\frac {1}{x} & =x+C\nonumber \\ v & =x^{2}+xC \tag {4} \end {align}
But \(v\relax (x) =y^{-3}\). Therefore the above becomes\begin {align*} y^{-3} & =x^{2}+xC\\ y^{3}\relax (x) & =\frac {1}{x^{2}+xC} \end {align*}
Or\[ y\relax (x) =\left (x^{2}+xC\right ) ^{-\frac {1}{3}}\]