Internal problem ID [8886]
Internal file name [OUTPUT/7821_Monday_June_06_2022_12_36_27_AM_66959053/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 552.
ODE order: 1.
ODE degree: 0.
The type(s) of ODE detected by this program : "first_order_nonlinear_p_but_separable"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
\[ \boxed {{y^{\prime }}^{n}-f \left (x \right ) g \left (y\right )=0} \]
The ode has the form \begin {align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end {align*}
Where \(n=n, m=1, f=f \left (x \right ) , g=g \left (y \right )\). Hence the ode is \begin {align*} (y')^{n} &= f \left (x \right ) g \left (y \right ) \end {align*}
Solving for \(y^{\prime }\) from (1) gives \begin {align*} y^{\prime } &={\mathrm e}^{\frac {\ln \left (f g \right )}{n}} \end {align*}
To be able to solve as separable ode, we have to now assume that \(f>0,g>0\). \begin {align*} f \left (x \right ) &> 0\\ g \left (y \right ) &> 0 \end {align*}
Under the above assumption the differential equations become separable and can be written as \begin {align*} y^{\prime } &=f^{\frac {1}{n}} g^{\frac {1}{n}} \end {align*}
Therefore \begin {align*} g^{-\frac {1}{n}} \, dy &= \left (f^{\frac {1}{n}}\right )\,dx \end {align*}
Replacing \(f(x),g(y)\) by their values gives \begin {align*} g \left (y \right )^{-\frac {1}{n}} \, dy &= \left (f \left (x \right )^{\frac {1}{n}}\right )\,dx \end {align*}
Integrating now gives the solutions. \begin {align*} \int g \left (y \right )^{-\frac {1}{n}}d y &= \int f \left (x \right )^{\frac {1}{n}}d x +c_{1} \end {align*}
Integrating gives \begin {align*} \int _{}^{y}g \left (y \right )^{-\frac {1}{n}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{n}}d x +c_{1} \end {align*}
Therefore \begin{align*} \int _{}^{y}g \left (y \right )^{-\frac {1}{n}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{n}}d x +c_{1} \\ \end{align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}g \left (y \right )^{-\frac {1}{n}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{n}}d x +c_{1} \\ \end{align*}
Verification of solutions
\[
\int _{}^{y}g \left (y \right )^{-\frac {1}{n}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{n}}d x +c_{1}
\] Verified OK. {0 < f(x), 0 < g(y)}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{n}-f \left (x \right ) g \left (y\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\frac {\ln \left (f \left (x \right ) g \left (y\right )\right )}{n}} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying exact <- exact successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 43
dsolve(diff(y(x),x)^n-f(x)*g(y(x))=0,y(x), singsol=all)
\[ \int _{}^{y \left (x \right )}g \left (\textit {\_a} \right )^{-\frac {1}{n}}d \textit {\_a} -g \left (y \left (x \right )\right )^{-\frac {1}{n}} \left (\int _{}^{x}\left (f \left (\textit {\_a} \right ) g \left (y \left (x \right )\right )\right )^{\frac {1}{n}}d \textit {\_a} \right )+c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 0.239 (sec). Leaf size: 41
DSolve[-(f[x]*g[y[x]]) + y'[x]^n==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}g(K[1])^{-1/n}dK[1]\&\right ]\left [\int _1^xf(K[2])^{\frac {1}{n}}dK[2]+c_1\right ] \]