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}