Internal
problem
ID
[9131] Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
linear
first
order Problem
number
:
150 Date
solved
:
Thursday, October 17, 2024 at 01:20:46 PM CAS
classification
:
[_linear]
Solve
\begin{align*} \left (x^{2}+1\right ) y^{\prime }+2 x y-2 x^{2}&=0 \end{align*}
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
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
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
Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the
function \(\phi \left (x,y\right )\)
Where \(f(x)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(x\) gives
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 the constant \(c_1\) gives the solution as
\[
c_1 = y \left (x^{2}+1\right )-\frac {2 x^{3}}{3}
\]
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)
\begin{align*} y = \frac {2 x^{3}+3 c_1}{3 x^{2}+3} \end{align*}
1.149.3 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+2 y \left (x \right ) x -2 x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate the derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {2 x y \left (x \right )}{x^{2}+1}+\frac {2 x^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+\frac {2 x y \left (x \right )}{x^{2}+1}=\frac {2 x^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (\frac {d}{d x}y \left (x \right )+\frac {2 x y \left (x \right )}{x^{2}+1}\right )=\frac {2 \mu \left (x \right ) x^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \left (x \right ) \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (\frac {d}{d x}y \left (x \right )+\frac {2 x y \left (x \right )}{x^{2}+1}\right )=\left (\frac {d}{d x}y \left (x \right )\right ) \mu \left (x \right )+y \left (x \right ) \left (\frac {d}{d x}\mu \left (x \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d x}\mu \left (x \right ) \\ {} & {} & \frac {d}{d x}\mu \left (x \right )=\frac {2 \mu \left (x \right ) x}{x^{2}+1} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=x^{2}+1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \left (x \right ) \mu \left (x \right )\right )\right )d x =\int \frac {2 \mu \left (x \right ) x^{2}}{x^{2}+1}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \left (x \right ) \mu \left (x \right )=\int \frac {2 \mu \left (x \right ) x^{2}}{x^{2}+1}d x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {\int \frac {2 \mu \left (x \right ) x^{2}}{x^{2}+1}d x +\mathit {C1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=x^{2}+1 \\ {} & {} & y \left (x \right )=\frac {\int 2 x^{2}d x +\mathit {C1}}{x^{2}+1} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\frac {2 x^{3}}{3}+\mathit {C1}}{x^{2}+1} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {2 x^{3}+3 \mathit {C1}}{3 x^{2}+3} \end {array} \]
1.149.4 Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order linear<-1st order linear successful`