8 L8: More on Exact + Reducible 2nd Order (2.6)
8.1 Recall
\[ Mdx + Ndy =0 \] is \(M_y = N_x\) ?
\[\boxed{\text{If }M_y=N_x,\text{ then,}\quad Mdx + Ndy =0\text{ is exact}}\]
Exact?
\[(3x+2y^2)dx+2xydy=0\]
\[M_y=4y,\qquad N_x=2y\] Not equal then Not exact! 😓
But, what if I multiply the whole equation by \(x\) ?
\[\Rightarrow\] \[(3x^2+2xy^2)dx+2x^2ydy=0\] \[\Rightarrow\]
\[M_y=4xy,\qquad N_x=4xy\] Becomes Exact!!! Awesome!!!
8.1.1 Now we can \(\int\):
\[ \frac{\partial F}{\partial x}=M \Rightarrow \int (3x^2+2xy^2)\,dx +\phi(y)\] \[\Rightarrow F= x^3+x^2y^2+\phi(y) \] \[ \frac{\partial F}{\partial y}= 0+x^22y+\phi'(y)=N=2x^2y\] \[\Rightarrow \phi'(y) = 0 \Rightarrow \phi(y) = C_1\]
Solution: \(F=C\)
\[\boxed{x^3+x^2y^2=C}\]
8.2 Definition: Integrating factor
A function \(\mu(x,y)\) for which \[\mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)dy=0\] is exact, is called an integrating factor
8.3 Theorem + proof
\[\mu Mdx+\mu Ndy=0\] \[\text{is exact } \Leftrightarrow \mu(M_y-N_x) = \mu_xN-\mu_yM\]
\[(\mu M)dx+(\mu N)dy=0\]
\[\text{exact }\Leftrightarrow \frac{\partial }{\partial y}(\mu M) =\frac{\partial }{\partial x}(\mu N) \]
\[ \Leftrightarrow \mu_yM + \mu M_y = \mu_x N + \mu N_x\] \[ \Leftrightarrow\mu(M_y-N_x)=\mu_xN-\mu_yM \] \[\square\]
Note:
\[\text{If } M_y-N_x=0, \quad\text{then}\quad \mu=1\quad\text{is a solution}\]
8.4 I.F. Separable? \(\mu(x,y)=P(x)Q(y)\) 1
\[\text{Suppose: }\boxed{\mu(x,y)=P(x)Q(y)}\]
Then \[\mu_x=P'(x)Q(y),\quad \mu_y=P(x)Q'(y)\]
and \[\mu(M_y-N_x)=\mu_xN-\mu_yM \] becomes
\[P(x)Q(y)(M_y-N_x)=P'(x)Q(y)N-P(x)Q'(y)M\] \[\Rightarrow M_y-N_x=\frac{P'(x)}{P(x)}N-\frac{Q'(y)}{Q(y)}M\]
Now let’s call:
\[p(x)=\frac{P'(x)}{P(x)},\qquad q(y)=\frac{Q'(y)}{Q(y)}\] becomes:
\[M_y-N_x=p(x)N-q(y)M\] Then now if we are able to find functions \(p(x)\) and \(q(x)\) which satisfy previous equation, we can said that the integrating factor \(\mu(x,y)\), can be written as a product of an \(x\)-function by a \(y\)-function.
8.5 Theorem on integrating factors
Let \(M,N,M_y,N_x\) be continuous on an open rectangle \(R\).
- \[\text{If }p(x)=\frac{M_y-N_x}{N}\text{ is independent of }y,\text{ then }\]
\[\boxed{\mu(x) =\pm e^{\int p(x)\,dx}}\] is an integrating factor for \(Mdx+Ndy=0\)
- \[\text{If }q(y)=\frac{N_x-M_y}{M}\text{ is independent of }x,\text{ then }\]
\[\boxed{\mu(y) =\pm e^{\int q(y)\,dy}}\] is an integrating factor for \(Mdx+Ndy=0\)
\[(2xy^3)dx + (3x^2y^2+x^2y^3+1)dy =0\]
\[M_y\stackrel{?}{=}N_x\] \[ 6xy^2 \neq 6xy^2+2xy^3\] \[\Rightarrow \text{try:}\] \[p=\frac{M_y-N_x}{N}\text{, is ind. of }y\quad ?\] \[\frac{6xy^2-(6xy^2+2xy^3)}{3x^2y^2+x^2y^3+1}=\frac{-2xy^3}{3x^2y^2+x^2y^3+1}\quad\text{, nope!}\]
\[\Rightarrow q=\frac{N_x-M_y}{M}\text{, is ind. of }x\quad ?\] \[\frac{6xy^2+2xy^3-6xy^2}{2xy^3}=\frac{2xy^3}{2xy^3}=1\text{, is ind. of }x\, !\] \[\Rightarrow \mu =e^{ \int q(y)dy}=e^y \] \[\Rightarrow e^y(2xy^3)dx + e^y(3x^2y^2+x^2y^3+1)dy =0 \] Then new M and N:
\[M=e^y(2xy^3)\qquad N=e^y(3x^2y^2+x^2y^3+1)\]
\[M=F_x\Rightarrow F=\int M\,dx + \phi(y)\] \[\Rightarrow F= 2y^3e^y\int xdx+\phi(y)=y^3e^yx^2+\phi(y)\] \[\Rightarrow F_y=x^2y^3e^y+x^23y^2e^y+\phi'(y)\] \[\Rightarrow N= e^y(3x^2y^2+x^2y^3+1)=x^2y^3e^y+x^23y^2e^y+\phi'(y)\]
\[\Rightarrow e^y=\phi'(y)\] \[\Rightarrow \phi(y)=e^y\] Final Solution2:
\[F=C\Leftrightarrow \boxed{y^3e^yx^2+e^y=C}\]
8.6 Reducible second order
A second-order differential equation involves the second derivative of the unknown function \(y(x)\), meaning it incorporates \(y''\), the rate at which the slope (\(y'\)) of the function \(y(x)\) changes with respect to \(x\). The general form of such an equation can include \(y\), \(y'\), \(y''\), and \(x\) in various combinations along with constants and possibly functions of \(x\).
8.6.1 Reducible Second-Order Equations
An equation is considered reducible if it can be simplified into a lower-order differential equation through a substitution. This process is particularly handy when either the dependent variable \(y\) or the independent variable \(x\) is missing from the equation, as it allows for a reduction to a first-order differential equation, which is often easier to solve.
8.6.1.1 Dependent Variable \(y\) Missing
When the dependent variable \(y\) is missing from a second-order differential equation, it means the equation only involves \(y'\), \(y''\), and possibly \(x\).
\[\boxed{F(y'',y',x)=0}\]3
Substitution Process:
- Define a New Variable for the First Derivative:
We set \[\boxed{p = \frac{dy}{dx}}\] meaning \(p\) represents the first derivative of \(y\) with respect to \(x\).
- Express the Second Derivative in Terms of the New Variable:
The second derivative of \(y\), which is \(y'' = \frac{d^2y}{dx^2}\), can be expressed in terms of \(p\) by noting that \[\boxed{y'' = \frac{dp}{dx}}\]
- Substitute into the Original Equation:
With this substitution, the original second-order equation, which involves \(y''\), transforms into a first-order differential equation in terms of \(p\) and \(x\).
This substitution effectively reduces the original second-order equation into a first-order equation involving \(p\) and \(x\), which can then be solved using standard methods for first-order differential equations. Once \(p(x)\) is found, it can be integrated to obtain \(y(x)\), since \(p = \frac{dy}{dx}\).
Solve the equation \[xy''+2y'=6x\]
\[p =\frac{dy}{dx}\rightarrow xy''+2y'=6x \rightarrow \boxed{xp'+2p=6x}\]
\[\cdots\]
8.7 References
Khan: Integrating factors 1 | First order differential equations https://youtu.be/j511hg7Hlbg
Khan: Integrating factors 2 | First order differential equations https://youtu.be/0NyeDUhKwBE