Internal problem ID [987]
Internal file name [OUTPUT/988_Sunday_June_05_2022_01_56_08_AM_70124900/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 2, First order equations. Transformation of Nonlinear Equations into Separable
Equations. Section 2.4 Page 68
Problem number: 9.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "bernoulli", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_homogeneous, `class G`], _rational, _Bernoulli]
\[ \boxed {x y^{\prime }+y-x^{4} y^{4}=0} \] With initial conditions \begin {align*} \left [y \left (1\right ) = {\frac {1}{2}}\right ] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {y \left (x^{4} y^{3}-1\right )}{x} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y={\frac {1}{2}}\) is \[
\{x <0\boldsymbol {\lor }0 The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y={\frac {1}{2}}\) is \[
\{x <0\boldsymbol {\lor }0
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \left (x^{4} y^{3}-1\right )}{x} \end {align*}
This is a Bernoulli ODE. \[ y' = -\frac {1}{x} y +x^{3} y^{4} \tag {1} \] The standard Bernoulli ODE has the form \[ y' = f_0(x)y+f_1(x)y^n \tag {2} \] The first step is to
divide the above equation by \(y^n \) which gives \[ \frac {y'}{y^n} = f_0(x) y^{1-n} +f_1(x) \tag {3} \] The next step is use the substitution \(w = y^{1-n}\) in equation
(3) which generates a new ODE in \(w \left (x \right )\) which will be linear and can be easily solved using an
integrating factor. Backsubstitution then gives the solution \(y(x)\) which is what we
want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows
that \begin {align*} f_0(x)&=-\frac {1}{x}\\ f_1(x)&=x^{3}\\ n &=4 \end {align*}
Dividing both sides of ODE (1) by \(y^n=y^{4}\) gives \begin {align*} y'\frac {1}{y^{4}} &= -\frac {1}{x \,y^{3}} +x^{3} \tag {4} \end {align*}
Let \begin {align*} w &= y^{1-n} \\ &= \frac {1}{y^{3}} \tag {5} \end {align*}
Taking derivative of equation (5) w.r.t \(x\) gives \begin {align*} w' &= -\frac {3}{y^{4}}y' \tag {6} \end {align*}
Substituting equations (5) and (6) into equation (4) gives \begin {align*} -\frac {w^{\prime }\left (x \right )}{3}&= -\frac {w \left (x \right )}{x}+x^{3}\\ w' &= \frac {3 w}{x}-3 x^{3} \tag {7} \end {align*}
The above now is a linear ODE in \(w \left (x \right )\) which is now solved.
Entering Linear first order ODE solver. In canonical form a linear first order is
\begin {align*} w^{\prime }\left (x \right ) + p(x)w \left (x \right ) &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {3}{x}\\ q(x) &=-3 x^{3} \end {align*}
Hence the ode is \begin {align*} w^{\prime }\left (x \right )-\frac {3 w \left (x \right )}{x} = -3 x^{3} \end {align*}
The integrating factor \(\mu \) is \begin{align*}
\mu &= {\mathrm e}^{\int -\frac {3}{x}d x} \\
&= \frac {1}{x^{3}} \\
\end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu w\right ) &= \left (\mu \right ) \left (-3 x^{3}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {w}{x^{3}}\right ) &= \left (\frac {1}{x^{3}}\right ) \left (-3 x^{3}\right )\\ \mathrm {d} \left (\frac {w}{x^{3}}\right ) &= -3\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {w}{x^{3}} &= \int {-3\,\mathrm {d} x}\\ \frac {w}{x^{3}} &= -3 x + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{x^{3}}\) results in \begin {align*} w \left (x \right ) &= c_{1} x^{3}-3 x^{4} \end {align*}
Replacing \(w\) in the above by \(\frac {1}{y^{3}}\) using equation (5) gives the final solution. \begin {align*} \frac {1}{y^{3}} = c_{1} x^{3}-3 x^{4} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y={\frac {1}{2}}\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 8 = c_{1} -3 \end {align*}
The solutions are \begin {align*} c_{1} = 11 \end {align*}
Trying the constant \begin {align*} c_{1} = 11 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {1}{y^{3}} = -3 x^{4}+11 x^{3} \end {align*}
The constant \(c_{1} = 11\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} \frac {1}{y^{3}} &= -3 x^{4}+11 x^{3} \\
\end{align*} Verification of solutions
\[
\frac {1}{y^{3}} = -3 x^{4}+11 x^{3}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x y^{\prime }+y-x^{4} y^{4}=0, y \left (1\right )=\frac {1}{2}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y+x^{4} y^{4}}{x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=\frac {1}{2} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 35
\[
y \left (x \right ) = \frac {\left (-\left (3 x -11\right )^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{6 x^{2}-22 x}
\]
✓ Solution by Mathematica
Time used: 0.415 (sec). Leaf size: 19
\[
y(x)\to \frac {1}{\sqrt [3]{-x^3 (3 x-11)}}
\]
5.13.2 Solving as bernoulli ode
5.13.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful`
dsolve([x*diff(y(x),x)+y(x)=x^4*y(x)^4,y(1) = 1/2],y(x), singsol=all)
DSolve[{x*y'[x]+y[x]==x^4*y[x]^4,y[1]==1/2},y[x],x,IncludeSingularSolutions -> True]