Internal problem ID [14217]
Internal file name [OUTPUT/13898_Saturday_March_09_2024_03_57_44_PM_39289556/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.2, page 39
Problem number: 58.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y^{\prime }+f \left (t \right ) y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= f( t) g(y)\\ &= -f \left (t \right ) y \end {align*}
Where \(f(t)=-f \left (t \right )\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= -f \left (t \right ) \,d t\\ \int { \frac {1}{y} \,dy} &= \int {-f \left (t \right ) \,d t}\\ \ln \left (y \right )&=\int -f \left (t \right )d t +c_{1}\\ y&={\mathrm e}^{\int -f \left (t \right )d t +c_{1}}\\ &=c_{1} {\mathrm e}^{\int -f \left (t \right )d t} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{1} {\mathrm e}^{\int _{}^{0}-f \left (\textit {\_a} \right )d \textit {\_a}} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=0 \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \end{align*}
Verification of solutions
\[ y = 0 \] Verified OK.
Entering Linear first order ODE solver. The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left ({\mathrm e}^{\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a}} y\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a}} y &= c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a}}\) results in \begin {align*} y &= c_{1} {\mathrm e}^{-\left (\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a} \right )} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=0 \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \end{align*}
Verification of solutions
\[ y = 0 \] Verified OK.
Using the change of variables \(y = u \left (t \right ) t\) on the above ode results in new ode in \(u \left (t \right )\) \begin {align*} u^{\prime }\left (t \right ) t +u \left (t \right )+f \left (t \right ) u \left (t \right ) t = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(t,u)\\ &= f( t) g(u)\\ &= -\frac {u \left (f \left (t \right ) t +1\right )}{t} \end {align*}
Where \(f(t)=-\frac {f \left (t \right ) t +1}{t}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {f \left (t \right ) t +1}{t} \,d t\\ \int { \frac {1}{u} \,du} &= \int {-\frac {f \left (t \right ) t +1}{t} \,d t}\\ \ln \left (u \right )&=\int -\frac {f \left (t \right ) t +1}{t}d t +c_{2}\\ u&={\mathrm e}^{\int -\frac {f \left (t \right ) t +1}{t}d t +c_{2}}\\ &=c_{2} {\mathrm e}^{\int -\frac {f \left (t \right ) t +1}{t}d t} \end {align*}
Therefore the solution \(y\) is \begin {align*} y&=t u\\ &=t c_{2} {\mathrm e}^{\int -\frac {f \left (t \right ) t +1}{t}d t} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(t=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = 0 \end {align*}
This solution is valid for any \(c_{2}\). Hence there are infinite number of solutions.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= t c_{2} {\mathrm e}^{\int -\frac {f \left (t \right ) t +1}{t}d t} \\ \end{align*}
Verification of solutions
\[ y = t c_{2} {\mathrm e}^{\int -\frac {f \left (t \right ) t +1}{t}d t} \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=-f \left (t \right ) y\\ y^{\prime }&= \omega \left ( t,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{t}+\omega \left ( \eta _{y}-\xi _{t}\right ) -\omega ^{2}\xi _{y}-\omega _{t}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is known. It is of type linear. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
|
ODE class |
Form |
\(\xi \) |
\(\eta \) |
|
linear ode |
\(y'=f(x) y(x) +g(x)\) |
\(0\) |
\(e^{\int fdx}\) |
|
separable ode |
\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) |
\(\frac {1}{f}\) |
\(0\) |
|
quadrature ode |
\(y^{\prime }=f\left ( x\right ) \) |
\(0\) |
\(1\) |
|
quadrature ode |
\(y^{\prime }=g\left ( y\right ) \) |
\(1\) |
\(0\) |
|
homogeneous ODEs of Class A |
\(y^{\prime }=f\left ( \frac {y}{x}\right ) \) |
\(x\) |
\(y\) |
|
homogeneous ODEs of Class C |
\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) |
\(1\) |
\(-\frac {b}{c}\) |
|
homogeneous class D |
\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) |
\(x^{2}\) |
\(xy\) |
|
First order special form ID 1 |
\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) |
\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
|
polynomial type ode |
\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) |
\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
|
Bernoulli ode |
\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\) |
\(0\) |
\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\) |
|
Reduced Riccati |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
|
|
|||
|
|
|||
The above table shows that \begin {align*} \xi \left (t,y\right ) &=0\\ \tag {A1} \eta \left (t,y\right ) &={\mathrm e}^{\int -f \left (t \right )d t} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( t,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d t}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial t} + \eta \frac {\partial }{\partial y}\right ) S(t,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = t \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{{\mathrm e}^{\int -f \left (t \right )d t}}} dy \end {align*}
Entering Exact first order ODE solver. (Form one type)
To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is \[ M(t,y) \mathop {\mathrm {d}t}+ N(t,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore \begin {align*} \left (-\frac {1}{y}\right )\mathop {\mathrm {d}y} &= \left (f \left (t \right )\right )\mathop {\mathrm {d}t}\\ \left (-f \left (t \right )\right )\mathop {\mathrm {d}t} + \left (-\frac {1}{y}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(t,y) &= -f \left (t \right )\\ N(t,y) &= -\frac {1}{y} \end {align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial t} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-f \left (t \right )\right )\\ &= 0 \end {align*}
And \begin {align*} \frac {\partial N}{\partial t} &= \frac {\partial }{\partial t} \left (-\frac {1}{y}\right )\\ &= 0 \end {align*}
Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial t}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (t,y\right )\) \begin {align*} \frac {\partial \phi }{\partial t } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end {align*}
Integrating (1) w.r.t. \(t\) gives \begin{align*} \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int M\mathop {\mathrm {d}t} \\ \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int -f \left (t \right )\mathop {\mathrm {d}t} \\ \tag{3} \phi &= \int _{0}^{t}-f \left (\textit {\_a} \right )d \textit {\_a}+ f(y) \\ \end{align*} Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(t\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives \begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = -\frac {1}{y}\). Therefore equation (4) becomes \begin{equation} \tag{5} -\frac {1}{y} = 0+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = -\frac {1}{y} \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {1}{y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\ln \left (y \right )+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \int _{0}^{t}-f \left (\textit {\_a} \right )d \textit {\_a} -\ln \left (y \right )+ c_{1} \] But since \(\phi \) itself is a constant function, then let \(\phi =c_{2}\) where \(c_{2}\) is new constant and combining \(c_{1}\) and \(c_{2}\) constants into new constant \(c_{1}\) gives the solution as \[ c_{1} = \int _{0}^{t}-f \left (\textit {\_a} \right )d \textit {\_a} -\ln \left (y \right ) \] The solution becomes\[ y = {\mathrm e}^{-\left (\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a} \right )-c_{1}} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = {\mathrm e}^{-c_{1}} \end {align*}
Unable to solve for constant of integration. Since \(\lim _{c_1 \to \infty }\) gives \(y = {\mathrm e}^{-\left (\int _{0}^{t}f \left (\textit {\_a} \right )d \textit {\_a} \right )-c_{1}}=y = 0\) and this result satisfies the given initial condition.
The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \end{align*}
Verification of solutions
\[ y = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+f \left (t \right ) y=0, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-f \left (t \right ) y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-f \left (t \right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int -f \left (t \right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\int -f \left (t \right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-\left (\int f \left (t \right )d t \right )+c_{1}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0={\mathrm e}^{-\left (\int ^{0}f \left (\textit {\_a} \right )d \textit {\_a} \right )+c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 5
dsolve([diff(y(t),t)+f(t)*y(t)=0,y(0) = 0],y(t), singsol=all)
\[ y \left (t \right ) = 0 \]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 6
DSolve[{y'[t]+f[t]*y[t]==0,{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to 0 \]