Find solution of the differential equation that satisfies the initial conditions \(\left ( 2xy+e^{y}\right ) dx+\left ( x^{2}+xe^{y}\right ) dy=0\) where \(y\left ( 1\right ) =\ln 2\)
Answer:
This is not separable, so we will try to see if it is exact. We write the ODE as\[ M\left ( x,y\right ) dx+N\left ( x,y\right ) dy=0 \] The condition for the ODE to be exact is \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\). Now \(\frac{\partial M}{\partial y}=2x+e^{y}\) and \(\frac{\partial N}{\partial x}=2x+e^{y}\), therefore it is exact. Let \begin{align} M & =\frac{\partial \varphi \left ( x,y\right ) }{\partial x}\tag{1}\\ N & =\frac{\partial \varphi \left ( x,y\right ) }{\partial y} \tag{2} \end{align}
Hence the ODE can be written as\begin{align*} \frac{\partial \varphi \left ( x,y\right ) }{\partial x}dx+\frac{\partial \varphi \left ( x,y\right ) }{\partial y}dy & =0\\ \frac{\partial \varphi \left ( x,y\right ) }{\partial x}+\frac{\partial \varphi \left ( x,y\right ) }{\partial y}\frac{dy}{dx} & =0\\ \frac{d}{dx}\left ( \varphi \left ( x,y\left ( x\right ) \right ) \right ) & =0 \end{align*}
This means that \(\varphi \left ( x,y\left ( x\right ) \right ) =C\), a constant. We need now to find \(\varphi \left ( x,y\right ) \). At this point we can pick either Eq. (1) or Eq. (2). Using Eq. (1) gives\[ 2xy+e^{y}=\frac{\partial \varphi \left ( x,y\right ) }{\partial x}\] Integrating\begin{align*} \varphi \left ( x,y\right ) & =\int 2xy+e^{y}dx\\ & =x^{2}y+xe^{y}+g\left ( y\right ) \end{align*}
Where \(g\left ( y\right ) \) acts here as the constant of integration, since \(y\left ( x\right ) \) is a function of \(x.\) Taking derivative of the above w.r.t. \(y\,\ \)and equating the result to Eq.(2) (or to \(N\left ( x,y\right ) \)) gives\begin{align*} x^{2}+xe^{y}+g^{\prime }\left ( y\right ) & =x^{2}+xe^{y}\\ g^{\prime }\left ( y\right ) & =0 \end{align*}
Hence \(g\left ( y\right ) \) is constant. We can choose any value for this constant, so we pick zero. Therefore\[ \varphi \left ( x,y\right ) =x^{2}y+xe^{y}\] But \(\varphi \left ( x,y\right ) =C\), hence\[ x^{2}y+xe^{y}=C \] What is left is to find \(C.\) For this we use the boundary conditions \(y\left ( 1\right ) =\ln 2\), which gives\begin{align*} 1^{2}\left ( \ln \left ( 2\right ) \right ) +1\times e^{\ln \left ( 2\right ) } & =C\\ C & =\ln \left ( 2\right ) +2 \end{align*}
Hence the solution is\[ x^{2}y+xe^{y}=2+\ln \left ( 2\right ) \] We can try to find explicit form for \(y\)\[ y+\frac{e^{y}}{x}=\frac{2+\ln \left ( 2\right ) }{x^{2}}\] Will leave it at this for now.
Solve the following differential equation \(2xy^{3}\left ( ydx+xdy\right ) =\left ( ydx-xdy\right ) \sin \left ( \frac{x}{y}\right ) \)
Answer:
Let us see if it is exact or not. Simplifying\begin{align*} 2xy^{4}dx+2x^{2}y^{3}dy & =\sin \left ( \frac{x}{y}\right ) ydx-x\sin \left ( \frac{x}{y}\right ) dy\\ \left ( 2xy^{4}-\sin \left ( \frac{x}{y}\right ) y\right ) dx+\left ( 2x^{2}y^{3}+x\sin \left ( \frac{x}{y}\right ) \right ) dy & =0 \end{align*}
Now we write the ODE as\[ M\left ( x,y\right ) dx+N\left ( x,y\right ) dy=0 \] Where\begin{align*} M\left ( x,y\right ) & =2xy^{4}-\sin \left ( \frac{x}{y}\right ) y\\ N\left ( x,y\right ) & =2x^{2}y^{3}+x\sin \left ( \frac{x}{y}\right ) \end{align*}
The condition for the ODE to be exact is \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\) \begin{align*} \frac{\partial M}{\partial y} & =8xy^{3}-\left ( \sin \left ( \frac{x}{y}\right ) +y\left ( \frac{-x}{y^{2}}\cos \left ( \frac{x}{y}\right ) \right ) \right ) \\ & =8xy^{3}-\sin \left ( \frac{x}{y}\right ) +\frac{x}{y}\cos \left ( \frac{x}{y}\right ) \end{align*}
and \[ \frac{\partial N}{\partial x}=4xy^{3}+\sin \left ( \frac{x}{y}\right ) +\frac{x}{y}\cos \left ( \frac{x}{y}\right ) \] Therefore it is not exact. To make it exact we need to use a general integrating factor. Trying \(I=\frac{1}{y^{2}}\). \(\ \)We multiply the ODE by this factor, which gives\begin{align*} M\left ( x,y\right ) & =\frac{1}{y^{2}}2xy^{4}-\frac{1}{y^{2}}\sin \left ( \frac{x}{y}\right ) y\\ & =2xy^{2}-\frac{1}{y}\sin \left ( \frac{x}{y}\right ) \\ N\left ( x,y\right ) & =\frac{1}{y^{2}}2x^{2}y^{3}+\frac{1}{y^{2}}x\sin \left ( \frac{x}{y}\right ) \\ & =2x^{2}y+\frac{x}{y^{2}}\sin \left ( \frac{x}{y}\right ) \end{align*}
Hence\begin{align*} \frac{\partial M}{\partial y} & =4xy-\left ( -\frac{1}{y^{2}}\sin \left ( \frac{x}{y}\right ) +\frac{1}{y}\left ( -\frac{x}{y^{2}}\right ) \cos \left ( \frac{x}{y}\right ) \right ) \\ & =4xy+\frac{1}{y^{2}}\sin \left ( \frac{x}{y}\right ) +\frac{x}{y^{3}}\cos \left ( \frac{x}{y}\right ) \end{align*}
And\begin{align*} \frac{\partial N}{\partial x} & =4xy+\frac{1}{y^{2}}\sin \left ( \frac{x}{y}\right ) +\frac{x}{y^{2}}\frac{1}{y}\cos \left ( \frac{x}{y}\right ) \\ & =4xy+\frac{1}{y^{2}}\sin \left ( \frac{x}{y}\right ) +\frac{x}{y^{3}}\cos \left ( \frac{x}{y}\right ) \end{align*}
Therefore it is exact now. Hence\begin{align*} M\left ( x,y\right ) & =2xy^{2}-\frac{1}{y}\sin \left ( \frac{x}{y}\right ) \\ N\left ( x,y\right ) & =2x^{2}y+\frac{x}{y^{2}}\sin \left ( \frac{x}{y}\right ) \end{align*}
Let \begin{align} M & =\frac{\partial \varphi \left ( x,y\right ) }{\partial x}\tag{1}\\ N & =\frac{\partial \varphi \left ( x,y\right ) }{\partial y} \tag{2} \end{align}
Hence the ODE can be written as\begin{align*} \frac{\partial \varphi \left ( x,y\right ) }{\partial x}dx+\frac{\partial \varphi \left ( x,y\right ) }{\partial y}dy & =0\\ \frac{\partial \varphi \left ( x,y\right ) }{\partial x}+\frac{\partial \varphi \left ( x,y\right ) }{\partial y}\frac{dy}{dx} & =0\\ \frac{d}{dx}\left ( \varphi \left ( x,y\left ( x\right ) \right ) \right ) & =0 \end{align*}
This means that \(\varphi \left ( x,y\left ( x\right ) \right ) =C\), a constant. We need now to find \(\varphi \left ( x,y\right ) \). At this point we can pick either Eq. (1) or Eq. (2). Using Eq. (1) gives\[ 2xy^{2}-\frac{1}{y}\sin \left ( \frac{x}{y}\right ) =\frac{\partial \varphi \left ( x,y\right ) }{\partial x}\] Integrating\begin{align*} \varphi \left ( x,y\right ) & =\int 2xy^{2}-\frac{1}{y}\sin \left ( \frac{x}{y}\right ) dx\\ & =x^{2}y^{2}+\frac{1}{y}y\cos \left ( \frac{x}{y}\right ) +g\left ( y\right ) \\ & =x^{2}y^{2}+\cos \left ( \frac{x}{y}\right ) +g\left ( y\right ) \end{align*}
Where \(g\left ( y\right ) \) acts here as the constant of integration, since \(y\left ( x\right ) \) is a function of \(x.\) Taking derivative of the above w.r.t. \(y\,\ \)and equating the result to Eq.(2) (or to \(N\left ( x,y\right ) \)) gives\begin{align*} 2yx^{2}-\left ( -\frac{x}{y^{2}}\right ) \sin \left ( \frac{x}{y}\right ) +g^{\prime }\left ( y\right ) & =2x^{2}y+\frac{x}{y^{2}}\sin \left ( \frac{x}{y}\right ) \\ 2yx^{2}+\frac{x}{y^{2}}\sin \left ( \frac{x}{y}\right ) +g^{\prime }\left ( y\right ) & =2x^{2}y+\frac{x}{y^{2}}\sin \left ( \frac{x}{y}\right ) \\ g^{\prime }\left ( y\right ) & =0 \end{align*}
Hence \(g\left ( y\right ) \) is constant. We can choose any value for this constant, so we pick zero. Therefore\[ \varphi \left ( x,y\right ) =x^{2}y^{2}+\cos \left ( \frac{x}{y}\right ) \] But \(\varphi \left ( x,y\right ) =C\), hence\[ x^{2}y^{2}+\cos \left ( \frac{x}{y}\right ) =C \] We are not given more information to find \(C\) so we stop here.
Find the solution which satisfies the given condition \(\left ( x^{4}+y^{4}\right ) dx=2x^{3}ydy\) where \(y\left ( 1\right ) =0\)
Solution:
This ODE is not separable, so we first check if it is exact. Writing the above as\begin{align*} \left ( x^{4}+y^{4}\right ) dx-2x^{3}ydy & =0\\ M\left ( x,y\right ) +N\left ( x,y\right ) \frac{dy}{dx} & =0 \end{align*}
Hence \(M\left ( x,y\right ) =\left ( x^{4}+y^{4}\right ) \) and \(N\left ( x,y\right ) =-2x^{3}y\). Lets check if it is exact first.\begin{align*} \frac{\partial M}{\partial y} & =4y^{3}\\ \frac{\partial N}{\partial x} & =-6yx^{2} \end{align*}
Therefore it is not exact. Finding the generalized integrating factor for the above was not easy. Using a small program and with the help of a CAS, the following integrating factor was found\[ I_{f}=\frac{1}{x\left ( y^{2}-x^{2}\right ) ^{2}}\] Hence the new \(M\left ( x,y\right ) \) is \(\frac{\left ( x^{4}+y^{4}\right ) }{x\left ( y^{2}-x^{2}\right ) ^{2}}\) and the new \(N\left ( x,y\right ) \) is \(\frac{-2x^{3}y}{x\left ( y^{2}-x^{2}\right ) ^{2}}.\) Therefore, now the following ODE is exact\[ \frac{\left ( x^{4}+y^{4}\right ) }{x\left ( y^{2}-x^{2}\right ) ^{2}}dx-\frac{2x^{3}y}{x\left ( y^{2}-x^{2}\right ) ^{2}}y=0 \] Let \begin{align} M & =\frac{\left ( x^{4}+y^{4}\right ) }{x\left ( y^{2}-x^{2}\right ) ^{2}}=\frac{\partial \varphi \left ( x,y\right ) }{\partial x}\tag{1}\\ N & =-\frac{2x^{3}y}{x\left ( y^{2}-x^{2}\right ) ^{2}}=\frac{\partial \varphi \left ( x,y\right ) }{\partial y} \tag{2} \end{align}
This means that \(\varphi \left ( x,y\left ( x\right ) \right ) =C\), a constant. We need now to find \(\varphi \left ( x,y\right ) \). At this point we can pick either Eq. (1) or Eq. (2). Using Eq. (1) gives\[ \frac{\left ( x^{4}+y^{4}\right ) }{x\left ( y^{2}-x^{2}\right ) ^{2}}=\frac{\partial \varphi \left ( x,y\right ) }{\partial x}\] Integrating\begin{align*} \varphi \left ( x,y\right ) & =\int \frac{\left ( x^{4}+y^{4}\right ) }{x\left ( y^{2}-x^{2}\right ) ^{2}}dx\\ & =\ln \left ( x\right ) -\frac{y^{2}}{x^{2}-y^{2}}+g\left ( y\right ) \end{align*}
Taking derivative of the above w.r.t. \(y\), and comparing the result to \(N\left ( x,y\right ) \) gives\[ \frac{-2y^{3}}{\left ( x^{2}-y^{2}\right ) ^{2}}-\frac{2y}{x^{2}-y^{2}}+g^{\prime }\left ( y\right ) =\frac{-2x^{3}y}{x\left ( y-x\right ) ^{2}\left ( y+x\right ) ^{2}}\] Simplifying\begin{align*} g^{\prime }\left ( y\right ) & =\frac{-2x^{3}y}{x\left ( y-x\right ) ^{2}\left ( y+x\right ) ^{2}}+\frac{2y^{3}}{\left ( x^{2}-y^{2}\right ) ^{2}}+\frac{2y}{x^{2}-y^{2}}\\ g^{\prime }\left ( y\right ) & =0 \end{align*}
Hence \(g\left ( y\right ) \) is constant. We can choose any value for this constant, so we pick zero. Therefore\[ \ln \left ( x\right ) -\frac{y^{2}}{x^{2}-y^{2}}=C \] is the solution. Using the condition \(y\left ( 1\right ) =0\) hence\begin{align*} \ln \left ( 1\right ) -\frac{0}{1-0} & =C\\ C & =0 \end{align*}
Hence the solution is\begin{align*} \ln \left ( x\right ) -\frac{y^{2}}{x^{2}-y^{2}} & =0\\ \left ( x^{2}-y^{2}\right ) \ln \left ( x\right ) -y^{2} & =0\\ x^{2}\ln \left ( x\right ) -y^{2}\ln \left ( x\right ) -y^{2} & =0\\ y^{2}\left ( 1+\ln \left ( x\right ) \right ) & =x^{2}\ln \left ( x\right ) \\ y^{2} & =\frac{x^{2}\ln \left ( x\right ) }{1+\ln \left ( x\right ) } \end{align*}
Therefore\[ y=\pm \frac{x\sqrt{\ln \left ( x\right ) }}{\sqrt{1+\ln \left ( x\right ) }} \]
Find the general solution for the equation, using the indicated solution for the homogeneous equation to reduce the order of the equation. \(\left ( 2x+1\right ) y^{\prime \prime }-4\left ( x+1\right ) y^{\prime }+4y=\frac{\left ( 2x+1\right ) ^{2}}{x+1}\)where \(y_{1}=e^{2x}\)
Solution:
Summary of method of solution: We let the second homogeneous solution be \(y_{2}=u\left ( x\right ) y_{1}\) and the substitute this back into the ODE. This gives a new ODE in \(u\) which we solve for \(u\). Once \(u\) is found, then homogeneous solution for the original ODE are found. Next we find the particular solution.
Let the second independent solution of the original ODE be\[ y_{2}\left ( x\right ) =u\left ( x\right ) y_{1}\left ( x\right ) \] Hence\begin{align*} y_{2}^{\prime } & =u^{\prime }y_{1}+uy_{1}^{\prime }\\ y_{2}^{\prime \prime } & =u^{\prime \prime }y_{1}+u^{\prime }y_{1}^{\prime }+u^{\prime }y_{1}^{\prime }+uy_{1}^{\prime \prime }\\ & =u^{\prime \prime }y_{1}+2u^{\prime }y_{1}^{\prime }+uy_{1}^{\prime \prime } \end{align*}
Substituting these back in the original ODE gives\begin{align*} \left ( 2x+1\right ) \left ( u^{\prime \prime }y_{1}+2u^{\prime }y_{1}^{\prime }+uy_{1}^{\prime \prime }\right ) -4\left ( x+1\right ) \left ( u^{\prime }y_{1}+uy_{1}^{\prime }\right ) +4\left ( uy_{1}\right ) & =0\\ \left ( 2x+1\right ) u^{\prime \prime }y_{1}+u^{\prime }\left ( \left ( 2x+1\right ) 2y_{1}^{\prime }-4\left ( x+1\right ) y_{1}\right ) +u\left ( \left ( 2x+1\right ) y_{1}^{\prime \prime }-4\left ( x+1\right ) y_{1}^{\prime }+4y_{1}\right ) & =0 \end{align*}
But \(\left ( 2x+1\right ) y_{1}^{\prime \prime }-4\left ( x+1\right ) y_{1}^{\prime }+4y_{1}\) in the last term above is the ODE itself, hence it is zero, therefore\[ \left ( 2x+1\right ) u^{\prime \prime }y_{1}+u^{\prime }\left ( \left ( 2x+1\right ) 2y_{1}^{\prime }-4\left ( x+1\right ) y_{1}\right ) =0 \] Now we substitute the actual value of \(y_{1}\left ( x\right ) \) \(\ \) and \(y_{1}^{\prime }\left ( x\right ) \) into the above\[ \left ( 2x+1\right ) u^{\prime \prime }e^{2x}+u^{\prime }\left ( \left ( 2x+1\right ) 4e^{2x}-4\left ( x+1\right ) e^{2x}\right ) =0 \] Dividing through by \(e^{2x}\) gives\begin{align*} \left ( 2x+1\right ) u^{\prime \prime }+4u^{\prime }\left ( \left ( 2x+1\right ) -\left ( x+1\right ) \right ) & =0\\ \left ( 2x+1\right ) u^{\prime \prime }+4xu^{\prime } & =0 \end{align*}
Let \(v=u^{\prime }\), hence\begin{align*} \left ( 2x+1\right ) v^{\prime }+4xv & =0\\ \frac{v^{\prime }}{v}+\frac{4x}{\left ( 2x+1\right ) } & =0\\ \ln v & =-\int \frac{4x}{\left ( 2x+1\right ) }dx\\ & =-\int 2-\frac{2}{2x+1}dx\\ & =-\int 2dx+2\int \frac{1}{2x+1}dx\\ & =-2x+\ln \left ( 2x+1\right ) +A \end{align*}
Hence\[ v=Ae^{-2x}\left ( 2x+1\right ) \] Where \(A\) is constant. Since only one second solution is needed, let \(A=1\). Therefore we have\begin{align*} \frac{du}{dx} & =e^{-2x}\left ( 2x+1\right ) \\ u & =\int e^{-2x}\left ( 2x+1\right ) dx\\ & =\int 2xe^{-2x}dx+\int e^{-2x}dx \end{align*}
Integration by parts is used for the first integral. The above simplifies to\[ u=-2e^{-2x}\left ( \frac{1}{4}+\frac{x}{2}\right ) +\frac{e^{-2x}}{-2}+B \] Where \(B\) is constant which can be set to zero. Hence \begin{align*} y_{2}\left ( x\right ) & =u\left ( x\right ) y_{1}\left ( x\right ) \\ & =e^{2x}\left ( -2e^{-2x}\left ( \frac{1}{4}+\frac{x}{2}\right ) +\frac{e^{-2x}}{-2}\right ) \\ & =-1-x \end{align*}
Therefore, the homogeneous is\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{2x}-c_{2}\left ( 1+x\right ) \\ & =c_{1}e^{2x}+c_{3}\left ( 1+x\right ) \end{align*}
Now we work on finding the particular solution. The forcing function is \(\frac{\left ( 2x+1\right ) ^{2}}{x+1}\). Using variation of parameters, since \(y_{1}=e^{2x},y_{2}=\left ( 1+x\right ) \) then \[ W\left ( x\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{2x} & \left ( 1+x\right ) \\ 2e^{2x} & 1 \end{vmatrix} =e^{2x}-2e^{2x}\left ( 1+x\right ) =-e^{2x}\left ( 1+2x\right ) \] Now we assume the particular solution is \[ y_{p}=u_{1}y_{1}+u_{2}y_{2}\] And recalling that the ODE is\[ \left ( 2x+1\right ) y^{\prime \prime }-4\left ( x+1\right ) y^{\prime }+4y=\frac{\left ( 2x+1\right ) ^{2}}{x+1}\] Hence\begin{align*} u_{1} & =\int \frac{-y_{2}f\left ( x\right ) }{W\left ( x\right ) a}dx=\int \frac{-\left ( 1+x\right ) \frac{\left ( 2x+1\right ) ^{2}}{x+1}}{-e^{2x}\left ( 1+2x\right ) \left [ 2x+1\right ] }dx=\int \frac{\left ( 2x+1\right ) }{e^{2x}\left ( 2x+1\right ) }dx=\int e^{-2x}dx\\ & =\frac{e^{-2x}}{-2} \end{align*}
Similarly, \begin{align*} u_{2} & =\int \frac{y_{1}f\left ( x\right ) }{W\left ( x\right ) a}dx=\int \frac{e^{2x}\frac{\left ( 2x+1\right ) ^{2}}{x+1}}{-e^{2x}\left ( 1+2x\right ) \left [ 2x+1\right ] }dx=-\int \frac{1}{x+1}dx\\ & =-\ln \left ( 1+x\right ) \end{align*}
Therefore, the particular solution is\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\frac{e^{-2x}}{-2}e^{2x}-\ln \left ( 1+x\right ) \left ( 1+x\right ) \\ & =-\frac{1}{2}-\ln \left ( 1+x\right ) \left ( 1+x\right ) \end{align*}
Hence the total solution is \(y=y_{h}+y_{p}\) or\[ y=c_{1}e^{2x}+c_{3}\left ( 1+x\right ) -\frac{1}{2}-\ln \left ( 1+x\right ) \left ( 1+x\right ) \] In expanded form\[ y=c_{1}e^{2x}+c_{3}+c_{3}x-\frac{1}{2}-\ln \left ( 1+x\right ) -x\ln \left ( 1+x\right ) \] The solution contains two constants of integration which can be found when given initial conditions.
Find the complete solution of \(\left ( D^{3}+D^{2}+3D-5\right ) y=e^{x}\)
Solution:
To find the roots of the characteristic equation, since it is a cubic polynomial, we guess the first root. we see that \(1\) is a root for \(\lambda ^{3}+\lambda ^{2}+3\lambda -5=0,\) now we do long division \(\left ( \lambda ^{3}+\lambda ^{2}+3\lambda -5\right ) /(\lambda -1)\) which gives \(\lambda ^{2}+2\lambda +5\), the roots of this are \(\frac{-b\pm \sqrt{b^{2}-4\left ( c\right ) }}{2}\)or \(\frac{-2\pm \sqrt{-16}}{2}\) or \(-1\pm 2i\), therefore the homogeneous solution is\[ y_{h}=c_{1}e^{x}+c_{2}e^{-1+2i}+c_{3}e^{-1-2i}\] Or in terms of \(\sin \) and \(\cos \) the above is\[ y_{h}=c_{1}e^{x}+e^{-x}\left ( A\cos \left ( 2x\right ) +B\sin \left ( 2x\right ) \right ) \] Now we need to find \(y_{p}\). Since \(e^{x}\) is part of \(y_{h}\) we have to guess \(y_{p}=c_{4}xe^{x}\). Substituting this into the original ODE gives\[ y_{p}^{\prime \prime \prime }+y_{p}^{\prime \prime }+3y_{p}^{\prime }-5y_{p}=e^{x}\] Now plugging the results of all the derivatives and dividing through by \(e^{x}\) gives\begin{align*} c_{4}\left ( 1+1+1+x\right ) +c_{4}\left ( 1+1+x\right ) +3c_{4}\left ( 1+x\right ) -5c_{4}x & =1\\ c_{4}\left ( 8\right ) +x\left ( c_{4}+c_{4}+3c_{4}-5c_{4}\right ) & =1\\ 8c_{4} & =1\\ c_{4} & =\frac{1}{8} \end{align*}
Therefore \[ y_{p}=\frac{1}{8}xe^{x}\] hence the full solution is \(y=y_{h}+y_{p}\) or\[ y=c_{1}e^{x}+e^{-x}\left ( A\cos \left ( 2x\right ) +B\sin \left ( 2x\right ) \right ) +\frac{1}{8}xe^{x}\]
Find the solution for the following equation \(3xy^{\prime }+y+x^{2}y^{4}=0\)
Solution:
Dividing by \(3x\) gives\[ y^{\prime }+\frac{1}{3x}y+\frac{x}{3}y^{4}=0 \] This is of the form \[ y^{\prime }+f\left ( x\right ) y+g\left ( x\right ) y^{n}=0 \] which is a Bernoulli ODE. Hence we start by dividing by \(y^{n}\) or \(y^{4}\) in this case, this gives\[ y^{-4}y^{\prime }+\frac{1}{3x}y^{-3}+\frac{x}{3}=0 \] Now let \(v\left ( x\right ) =y^{-3}\), therefore \(v^{\prime }\left ( x\right ) =-3y^{-4}y^{\prime }\left ( x\right ) \). Now we substitute these into the above ODE which turns it to an ODE in \(v\) that we can solve\begin{align*} \frac{-1}{3}v^{\prime }+\frac{1}{3x}v+\frac{x}{3} & =0\\ v^{\prime }-\frac{1}{x}v & =x \end{align*}
Hence \[ v\left ( x\right ) =\left ( x+c\right ) x \] or\begin{align*} \frac{1}{y^{3}} & =\left ( x+c\right ) x\\ y^{3} & =\frac{1}{\left ( x+c\right ) x}\\ y & =\left ( \frac{1}{\left ( x+c\right ) x}\right ) ^{\frac{1}{3}} \end{align*}
Solve \(\left ( y^{\prime }\right ) ^{2}y^{\prime \prime }=1+\left ( y^{\prime }\right ) ^{2}\)
Solution
We start by writing the ODE as\[ y^{\prime \prime }=\overset{f\left ( x,y,y^{\prime }\right ) }{\overbrace{\left ( y^{\prime }\right ) ^{-2}+1}}\] Therefore, \(f\left ( x,y,y^{\prime }\right ) =1+\left ( y^{\prime }\right ) ^{2}\) So we see that \(f\) is missing both \(x\) and \(y\). i.e. \(y^{\prime \prime }=f\left ( y^{\prime }\right ) \). So we will try both cases covered in class that handle missing \(x\) and missing \(y\) and see which produces a solution.
Let \(u\left ( x\right ) =y^{\prime }\left ( x\right ) \), hence \(u^{\prime }=y^{\prime \prime }\) and the ODE becomes\begin{align*} u^{2}u^{\prime } & =1+u^{2}\\ \frac{du}{dx} & =\frac{1}{u^{2}}+1 \end{align*}
This is now separable\begin{align*} \frac{du}{\frac{1}{u^{2}}+1} & =dx\\ \int \frac{u^{2}}{1+u^{2}}du & =\int dx\\ \int 1-\frac{1}{u^{2}+1}du & =x+c\\ u-\arctan \left ( u\right ) & =x+c \end{align*}
Since \(u\left ( x\right ) =y^{\prime }\left ( x\right ) \), the above becomes \[ y^{\prime }-\arctan \left ( y^{\prime }\right ) =x+c \] This is an implicit solution for \(y\left ( x\right ) .\)
Let \(v\left ( y\right ) =y^{\prime }\left ( x\right ) \), but we notice now that \(v\left ( y\right ) \) is function of \(y\). Hence \(y^{\prime \prime }=v\frac{dv}{dy}\) and the ODE becomes\[ v\frac{dv}{dy}=\frac{1}{v^{2}}+1 \] This is now separable\begin{align*} \frac{v}{\frac{1}{v^{2}}+1}dv & =dy\\ \int \frac{v^{3}}{1+v^{2}}dv & =\int dy\\ \int \frac{v^{3}}{1+v^{2}}dv & =\int dy\\ \int v-\frac{v}{v^{2}+1}dv & =y+c\\ \frac{v^{2}}{2}-\frac{1}{2}\ln \left ( 1+v^{2}\right ) & =y+c \end{align*}
Since \(v=y^{\prime }\left ( x\right ) \), the above becomes \[ \left ( y^{\prime }\right ) ^{2}-\ln \left ( 1+\left ( y^{\prime }\right ) ^{2}\right ) =2y+c_{1}\] This is an implicit solution for \(y\left ( x\right ) .\)
Obtain the solutions of the simultaneous equations\begin{align*} x^{\prime }+y^{\prime }+x & =-e^{-t}\\ x^{\prime }+2y^{\prime }+2x+2y & =0 \end{align*}
with initial conditions \(x\left ( 0\right ) =-1;y\left ( 0\right ) =1\)
Solution:
Taking the Laplace transform of the above system of equations, and using \(X\equiv \mathcal{L}\left ( x\left ( t\right ) \right ) \) and \(Y=\mathcal{L}\left ( y\left ( t\right ) \right ) \) gives\begin{align*} sX-x\left ( 0\right ) +sY-y\left ( 0\right ) +X & =-\frac{1}{s+1}\\ sX-x\left ( 0\right ) +2\left ( sY-y\left ( 0\right ) \right ) +2X+2Y & =0 \end{align*}
Substituting initial conditions gives\begin{align*} sX+1+sY-1+X & =-\frac{1}{s+1}\\ sX+1+2\left ( sY-1\right ) +2X+2Y & =0 \end{align*}
Simplifying\begin{align*} X\left ( 1+s\right ) +sY & =-\frac{1}{s+1}\\ X\left ( s+2\right ) +Y\left ( 2s+2\right ) & =1 \end{align*}
Hence\[\begin{pmatrix} 1+s & s\\ s+2 & 2\left ( s+1\right ) \end{pmatrix}\begin{pmatrix} X\\ Y \end{pmatrix} =\begin{pmatrix} -\frac{1}{s+1}\\ 1 \end{pmatrix} \] Therefore\begin{align*} \begin{pmatrix} X\\ Y \end{pmatrix} & =\begin{pmatrix} 1+s & s\\ s+2 & 2\left ( s+1\right ) \end{pmatrix} ^{-1}\begin{pmatrix} -\frac{1}{s+1}\\ 1 \end{pmatrix} \\ & =\frac{1}{s^{2}+2s+2}\begin{pmatrix} 2s+2 & -s\\ -s-2 & s+1 \end{pmatrix}\begin{pmatrix} -\frac{1}{s+1}\\ 1 \end{pmatrix} \\ & =\frac{1}{s^{2}+2s+2}\begin{pmatrix} \left ( 2s+2\right ) \left ( -\frac{1}{s+1}\right ) -s\\ \left ( -s-2\right ) \left ( -\frac{1}{s+1}\right ) +s+1 \end{pmatrix} \\ & =\frac{1}{s^{2}+2s+2}\begin{pmatrix} -s-\frac{2s+2}{s+1}\\ s+\frac{1}{s+1}\left ( s+2\right ) +1 \end{pmatrix} \\ & =\begin{pmatrix} -\frac{s+\frac{2s+2}{s+1}}{s^{2}+2s+2}\\ \frac{s+\frac{1}{s+1}\left ( s+2\right ) +1}{s^{2}+2s+2}\end{pmatrix} \end{align*}
Hence \begin{align*} X & =-\frac{s+\frac{2s+2}{s+1}}{s^{2}+2s+2}=-\frac{s+2}{s^{2}+2s+2}\\ Y & =\frac{s+\frac{1}{s+1}\left ( s+2\right ) +1}{s^{2}+2s+2}=\frac{\left ( s^{2}+3s+3\right ) }{\left ( s+1\right ) \left ( s^{2}+2s+2\right ) } \end{align*}
Now the inverse Laplace transform of each is found. Looking at \(X\)\begin{align*} X & =-\frac{s}{s^{2}+2s+2}-\frac{2}{s^{2}+2s+2}\\ & =-\frac{s}{\left ( s-a\right ) \left ( s-b\right ) }-2\frac{1}{\left ( s-a\right ) \left ( s-b\right ) } \end{align*}
Where \(a=-1-i,b=-1+i\). Now from tables, \(\mathcal{L}^{-1}\left ( \frac{s}{\left ( s-a\right ) \left ( s-b\right ) }\right ) =\frac{1}{a-b}\left ( ae^{at}-be^{bt}\right ) \), and \(\mathcal{L}^{-1}\left ( \frac{1}{\left ( s-a\right ) \left ( s-b\right ) }\right ) =\frac{1}{a-b}\left ( e^{at}-e^{bt}\right ) \), hence\begin{align*} x\left ( t\right ) & =\frac{-1}{a-b}\left ( ae^{at}-be^{bt}\right ) -2\frac{1}{a-b}\left ( e^{at}-e^{bt}\right ) \\ & =\frac{-1}{\left ( -1-i\right ) -\left ( -1+i\right ) }\left ( \left ( -1-i\right ) e^{at}-\left ( -1+i\right ) e^{bt}\right ) \\ & -2\frac{1}{\left ( -1-i\right ) -\left ( -1+i\right ) }\left ( e^{at}-e^{bt}\right ) \\ & \\ & =\frac{1}{2i}\left ( -e^{at}-ie^{at}+e^{bt}-ie^{bt}\right ) -2\frac{1}{-2i}\left ( e^{at}-e^{bt}\right ) \\ & =-\frac{1}{2i}e^{at}-\frac{1}{2i}ie^{at}+\frac{1}{2i}e^{bt}-\frac{1}{2i}ie^{bt}+\frac{1}{i}e^{at}-\frac{1}{i}e^{bt}\\ & =\frac{1}{2i}e^{at}-\frac{1}{2i}e^{bt}-\frac{1}{2i}ie^{at}-\frac{1}{2i}ie^{bt} \end{align*}
Now substitute the values for \(a\) and \(b\) in the exponents\begin{align*} x\left ( t\right ) & =\frac{1}{2i}e^{\left ( -1-i\right ) t}-\frac{1}{2i}e^{\left ( -1+i\right ) t}-\frac{1}{2i}ie^{\left ( -1-i\right ) t}-\frac{1}{2i}ie^{\left ( -1+i\right ) t}\\ & =\frac{1}{2i}\left ( e^{-t}e^{-it}\right ) -\frac{1}{2i}\left ( e^{-t}e^{it}\right ) -\frac{1}{2}\left ( e^{-t}e^{-it}\right ) -\frac{1}{2}\left ( e^{-t}e^{it}\right ) \\ & =e^{-t}\left ( \frac{e^{-it}-e^{it}}{2i}\right ) -e^{-t}\left ( \frac{e^{-it}+e^{it}}{2}\right ) \\ & =e^{-t}\left ( \frac{-ie^{-it}+ie^{it}}{2}\right ) -e^{-t}\cos t\\ & =-e^{-t}\left ( \sin t\right ) -e^{-t}\cos t\\ & =-e^{-t}\left ( \sin t+\cos t\right ) \end{align*}
Now to find the inverse Laplace of \(Y\)\begin{align*} Y & =\frac{\left ( s^{2}+3s+3\right ) }{\left ( s+1\right ) \left ( s^{2}+2s+2\right ) }\\ & =\frac{1}{s^{2}+2s+2}+\frac{1}{s+1}\\ & =\frac{1}{\left ( s-a\right ) \left ( s-b\right ) }+\frac{1}{s+1} \end{align*}
Where \(a=-1-i,b=-1+i\). But \(\mathcal{L}^{-1}\left ( \frac{1}{\left ( s-a\right ) \left ( s-b\right ) }\right ) =\frac{1}{a-b}\left ( e^{at}-e^{bt}\right ) =\frac{1}{-2i}\left ( e^{\left ( -1-i\right ) t}-e^{\left ( -1-i\right ) t}\right ) \), Hence\begin{align*} \mathcal{L}^{-1}\left ( \frac{1}{\left ( s-a\right ) \left ( s-b\right ) }\right ) & =\frac{1}{-2i}\left ( e^{-t}e^{-it}-e^{-t}e^{-it}\right ) \\ & =-e^{-t}\left ( \frac{-ie^{-it}+ie^{-it}}{2}\right ) \\ & =e^{-t}\left ( \frac{ie^{-it}-ie^{-it}}{2}\right ) \\ & =e^{-t}\sin t \end{align*}
And \(\mathcal{L}^{-1}\left ( \frac{1}{s+1}\right ) =e^{-t}\), hence \[ y\left ( t\right ) =e^{-t}\sin t+e^{-t}\]