Calculus 2 - First Order Differential Equation

Contents: Integral Infinite Series First Order Differential Equation Introduction Solving ODE Reference: Paul's Online Notes: ODE

Introduction

Definition
When the unknown variable function depends on a single independent variable, only ordinary derivatives appear in the equation. So, this equation is said to be an Ordinary Differential Equation (ODE).

The order of a differential equation is the order of the highest derivative in the equation.

The general linear ODE has the form:
$a_0(t)y^{(n)} + a_1(t)y^{(n-1)} + ... + a_n(t)y = g(t)$

A linear first order ODE has the general form:
$\frac{dy}{dt} + p(t)y = g(t)$

  Example 6.1 - Linear ODEs

Identify whether the given first ODE is a linear first ODE or not.
$\begin{align*} \text{A. } &y' + y = x \\ \text{B. } &y\frac{dy}{dt}+2y = e^t\\ \text{C. } &\frac{dy}{dx} + \cos{y} = x^3 \\ \text{D. } &t\frac{dy}{dt}-y-t=0 \\ \text{E. } &y\cos{y} + \frac{1}{x}\frac{dy}{dx} - \ln{x} = 0 \\ \end{align*}$

Solution:
Basically, a first ODE will be linear if it can be expressed in the standard form:
$\frac{dy}{dt} + p(t)y = g(t)$

1. So:
   $\begin{align*} y' + y &= x \\ p(x) &= 1\\ g(x) &= x \end{align*}$
   
   
   Since it CAN be written in the standard form, it is Linear.


2. So:
   $\begin{align*} y\frac{dy}{dt}+2y &= e^t \\ \frac{dy}{dt}+2 &= \frac{e^t}{y} \end{align*}$
   
   
   Since it CANNOT be written in the standard form, it is Not Linear.


3. So:
   $\begin{align*} \frac{dy}{dx} + \cos{y} = x^3 \end{align*}$
   
   
   Since it CANNOT be written in the standard form, it is Not Linear.


4. So:
   $\begin{align*} t\frac{dy}{dt}-y-t&=0\\ \frac{dy}{dt}-\frac{y}{t}-1&=0\\ \frac{dy}{dt}-\frac{y}{t}&=1\\ \\ p(t) &= -\frac{1}{t}\\ g(t) &= 1 \end{align*}$
   
   
   Since it CAN be written in the standard form, it is Linear.


5. So:
   $\begin{align*} y\cos{y} + \frac{1}{x}\frac{dy}{dx} - \ln{x} &= 0\\ xy\cos{y} + \frac{dy}{dx} - x\ln{x} &= 0\\ \frac{dy}{dx} + xy\cos{y} &= x\ln{x}\\ \end{align*}$
   
   
   Since it CANNOT be written in the standard form, it is Not Linear.

Solving First ODE

Separable ODEs
Separable equation usually has the form:
$M(x) \; dx + N(y) \; dy = 0$

We can solve the equation above directly:
$\begin{align*} M(x) \; dx + N(y) \; dy &= 0\\ M(x) \; dx &= -N(y) \; dy\\ \int M(x) \; dx &= - \int N(y) \; dy\\ \end{align*}$

  Example 6.2 - Separable Method I

Identify whether the given first ODE is a separable first ODE or not.
$\begin{align*} \text{A. } &(x^2-y^2)y' + xy = 0\\ \text{B. } &(x^2y^2+y^2)y' + x = 0\\ \text{C. } &(x\sin{y} + x^2)y' + \cos{x} = 0\\ \text{D. } &\frac{dy}{dx}=e^{y-x} .\sec{y}.(1+x^2)\\ \text{E. } &(x\sin{y}+xy)y' + (x^2+1)y = 0 \\ \end{align*}$

Solution:
Basically, a first ODE is a separable first ODE if it can be manipulated into the following form:
$M(x)\;dx = N(y) \; dy$

1. So:
   $\begin{align*} (x^2-y^2)y' + xy &= 0\\ (x^2-y^2)\frac{dy}{dx}&= -xy\\ \end{align*}$
   
   
   This ODE cannot be manipulated further to separate x and y, due to the lack of common factor. Hence, it is Not Separable.


2. So:
   $\begin{align*} (x^2y^2+y^2)y' + x &= 0 \\ (x^2y^2+y^2)\frac{dy}{dx} &= -x\\ (x^2+1)y^2\frac{dy}{dx} &= -x\\ y^2\;dy &= -\frac{x}{x^2+1} \;dx\\ \end{align*}$
   
   
   Hence, it is Separable.


3. So:
   $\begin{align*} (x\sin{y} + x^2)y' + \cos{x} &= 0\\ (x\sin{y} + x^2)\frac{dy}{dx} + \cos{x} &= 0\\ (x\sin{y} + x^2)\frac{dy}{dx} &= -\cos{x}\\ (\sin{y} + x)\frac{dy}{dx} &= -\frac{\cos{x}}{x}\\ \end{align*}$
   
   
   This ODE cannot be manipulated further to separate x and y, due to the lack of common factor. Hence, it is Not Separable.


4. So:
   $\begin{align*} \frac{dy}{dx}&=e^{y-x} .\sec{y}.(1+x^2)\\ \frac{dy}{dx}&=\frac{e^y}{e^x} .\sec{y}.(1+x^2)\\ \frac{1}{e^y \sec{y}}\;dy&=\frac{1+x^2}{e^x}\;dx\\ \end{align*}$
   
   
   Hence, it is Separable.


5. So:
   $\begin{align*} (x\sin{y}+xy)y' + (x^2+1)y &= 0 \\ (x\sin{y}+xy)\frac{dy}{dx} + (x^2+1)y &= 0 \\ (\sin{y}+y)x\frac{dy}{dx} + (x^2+1)y &= 0 \\ (\sin{y}+y)x\frac{dy}{dx} &= - (x^2+1)y \\ \frac{\sin{y}+y}{y} \;dy&= -\frac{x^2+1}{x} \;dx \end{align*}$
   
   
   Hence, it is Separable.

  Example 6.3 - Separable Method II

Find the solution of the given first ODE and initial condition:
$\begin{align*} \text{A. } &(x^2y^2+y^2)y'+x=0 \\ &y(0)=0\\ \text{B. } & y' = \frac{3x^2+4x-4}{2y-4} \\ &y(1)=3\\ \text{C. } &\frac{dy}{dt}=e^{y-t} .\sec{y}.(1+t^2) \\ &y(0)=0\\ \end{align*}$

Solution:

1. Separating variables:
   $\begin{align*} (x^2y^2+y^2)y'+x &=0 \\ (x^2y^2+y^2)y'&=-x \\ (x^2+1)(y^2)(y')&=-x\\ y^2 \;dy &= -\frac{x}{x^2+1}\;dx \end{align*}$
   
   The right hand side integral requires the substitution method, which is shown below:
   $\begin{align*} u&=x^2+1\\ du&=2x \; dx\\ \frac{du}{2} &= x\;dx\\ \\ \int -\frac{x}{x^2+1}\;dx&= \int - \frac{1}{2u}\;du\\ &= -\frac{1}{2}\ln{|u|}+C\\ &= -\frac{1}{2}\ln{|x^2+1|}+C \end{align*}$
   
   Now, we return to our main problem:
   $\begin{align*} y^2 \;dy &= -\frac{x}{x^2+1}\;dx\\ \int y^2 \;dy &= \int -\frac{x}{x^2+1}\;dx\\ \frac{1}{3}y^3 &= -\frac{1}{2}\ln{|x^2+1|}+C\\ \end{align*}$
   
   Using the initial condition, we can find the exact value of C:
   $\begin{align*} \frac{1}{3}y^3 &= -\frac{1}{2}\ln{|x^2+1|}+C\\ C &= 0\\ \end{align*}$
   
   This is our final solution:
   $\begin{align*} \frac{1}{3}y^3 &= -\frac{1}{2}\ln{|x^2+1|}\\ y&= \sqrt[3]{\frac{3}{2}\ln{|x^2+1|}} \\ \end{align*}$


2. Separating variables:
   $\begin{align*} y' &= \frac{3x^2+4x-4}{2y-4} \\ \frac{dy}{dx} &= \frac{3x^2+4x-4}{2y-4}\\ (2y-4)\;dy &= (3x^2+4x-4)\;dx\\ \end{align*}$
   
   Since both integration are simple (as in not requiring special methods), I will directly do both integrals:
   $\begin{align*} (2y-4)\;dy &= (3x^2+4x-4)\;dx\\ \int (2y-4)\;dy &= \int (3x^2+4x-4)\;dx\\ y^2 - 4y &= x^3 + 2x^2 -4x + C\\ \end{align*}$
   
   Using the initial condition, we can find the exact value of C:
   $\begin{align*} y^2 - 4y &= x^3 + 2x^2 -4x + C\\ 9 - 12 &= 1 +2 -4 + C\\ C &= -2 \end{align*}$
   
   This can be our final solution:
   $y^2 - 4y = x^3 + 2x^2 -4x -2\\ $
   
   However, we can express this solution in explicit form. Notice how when we manipulate it into this, it becomes a quadratic equation where we can solve for y:
   $\begin{align*} y^2 - 4y &= x^3 + 2x^2 -4x -2\\ y^2-4y-(x^3+2x^2-4x-2) &= 0\\ \\ y&=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ &=\frac{4 \pm \sqrt{16+4(x^3+2x^2-4x-2)}}{2}\\ &=2 \pm \sqrt{4+(x^3+2x^2-4x-2)}\\ &=2 \pm \sqrt{x^3+2x^2-4x+2)}\\ \end{align*}$
   
   The plus/minus sign implies that there are multiple solutions (of the quadratic equation), when there is only one correct solution (for this entire problem). We should check whether the solution is correct with the + or - sign from the initial condition. In this case, our final solution will be using the + sign since the initial condition would still be true:
   $y=2 + \sqrt{x^3+2x^2-4x+2)}$


3. Separating variables:
   $\begin{align*} \frac{dy}{dt}&=e^{y-t}\sec{y}(1+t^2)\\ \frac{dy}{dt}&=\frac{e^y}{e^t}\sec{y}(1+t^2)\\ \frac{\cos{y}}{e^y} \; dy &= \frac{1+t^2}{e^t}\;dt\\ \end{align*}$
   
   Integrating left hand side (Two integration by parts will be done):
   $\begin{align*} u &= \cos{y}\\ du &= -\sin{y}\;dy\\ dv &= e^{-y}\;dy\\ v &= -e^{-y}\\ A&=\int \frac{\cos{y}}{e^y} \; dy \\ &= uv - \int v \; du\\ &= -e^{-y} \cos{y} - \int e^{-y}\sin{y}\;dy\\ \\ a &= \sin{y}\\ da &= \cos{y}\;dy\\ db &= e^{-y}\;dy\\ b &= -e^{-y}\\ \\ A&= -e^{-y} \cos{y} - \int e^{-y}\sin{y}\;dy\\ &= -e^{-y} \cos{y} - (ab - \int b\;da)\\ &= -e^{-y} \cos{y} - (-e^{-y}\sin{y} + \int e^{-y}\cos{y}\;dy)\\ &= -e^{-y} \cos{y} - (-e^{-y}\sin{y} + A)\\ 2A &= e^{-y}\sin{y} - e^{-y}\cos{y}\\ A&= \frac{e^{-y}}{2} (\sin{y} - \cos{y}) + C_A \end{align*}$
   
   Now, the right hand side (The method used is similar to integration by parts):
   

   Derivative	Integral
   1+t2	e-t
   2t	-e-t
   2	e-t
   0	-e-t

   
   
   The result of derivation and integration in same colored cells are multiplied, and the sign between terms alternate:
   $\begin{align*} B&=\int \frac{1+t^2}{e^t}\;dt\\ &=\text{Yellow} - \text{Orange} + \text{Blue}\\ &=(1+t^2)(-e^{-t}) - (2t)(e^{-t}) + (-2e^{-t})+C_B\\ &=(-e^{-t})(t^2+2t+3)+C_B\\ \end{align*}$
   
   Now, returning to the main problem:
   $\begin{align*} \frac{dy}{dt}&=e^{y-t}\sec{y}(1+t^2)\\ \frac{dy}{dt}&=\frac{e^y}{e^t}\sec{y}(1+t^2)\\ \frac{\cos{y}}{e^y} \; dy &= \frac{1+t^2}{e^t}\;dt\\ \frac{e^{-y}\sin{y} - e^{-y}\cos{y}}{2} &= (-e^{-t})(t^2+2t+3)+C\\ \frac{e^{-y}}{2} (\sin{y} - \cos{y}) &= (-e^{-t})(t^2+2t+3)+C\\ \end{align*}$
   
   Using the initial condition, we can find the exact value of C:
   $\begin{align*} \frac{e^{-y}}{2} (\sin{y} - \cos{y}) &= (-e^{-t})(t^2+2t+3)+C\\ \frac{1}{2} (0-1) &= (-1)(3)+C\\ C &= \frac{5}{2} \end{align*}$
   
   This is our final solution:
   $\frac{e^{-y}}{2} (\sin{y} - \cos{y}) = (-e^{-t})(t^2+2t+3)+\frac{5}{2}\\ $

Variable Coefficients
Suppose we have a first order linear equation:
$\frac{dy}{dt} + p(t)y = g(t)$

We can use the Integrating Factor shown below:
$\mu (t) = e^{\int p(t) \; dt}$

Using that, we can find the solution of the ODE:
$y = \frac{1}{\mu} \left ( \int \mu.g \;dt + C \right )\\ $

  Example 6.4 - Variable Coefficients

Find the general solution of the given ODE:
$\begin{align*} &\text{A. } ty' + 2y = t^2 - t + 1 \\ &\text{B. } y' + 2y = e^{\frac{t}{2}} \\ &\text{C. } \frac{y'}{t} - \frac{2}{t^2}y = t \cos{t} \\ \end{align*}$

Solution:

1. Determining p and g:
   $\begin{align*} ty' + 2y &= t^2 - t + 1\\ y' + \frac{2}{t}y &= t - 1 + \frac{1}{t}\\ \\ p(t) &= \frac{2}{t}\\ g(t) &= t - 1 + \frac{1}{t}\\ \\ \mu (t) &= e^{\int p(t) \; dt}\\ &= e^{\int \frac{2}{t} \; dt}\\ &= e^{\ln{t}^2}\\ &= t^2 \\ y&= \frac{1}{\mu} \left ( \int \mu.g \;dt + C \right )\\ &= \frac{1}{t^2} \left ( \int (t^3 - t^2 + t) \;dt + C \right )\\ &= \frac{1}{t^2} \left ( \frac{1}{4} t^4 - \frac{1}{3} t^3 + \frac{1}{2} t^2 + C \right )\\ \end{align*}$


2. Determining p and g:
   $\begin{align*} y' + 2y = e^{\frac{t}{2}}\\ \\ p(t) &= 2\\ g(t) &= e^{\frac{t}{2}}\\ \\ \mu (t) &= e^{\int p(t) \; dt}\\ &= e^{\int \frac{2} \; dt}\\ &= e^{2t}\\ \\ y&= \frac{1}{\mu} \left ( \int \mu.g \;dt + C \right )\\ &= \frac{1}{e^{2t}} \left ( \int e^{\frac{5}{2}t} \;dt + C \right )\\ &= \frac{1}{e^{2t}} \left ( \frac{2}{5} e^{\frac{5}{2}t}+ C \right )\\ \end{align*}$


3. Determining p and g:
   $\begin{align*} \frac{y'}{t} - \frac{2}{t^2}y& = t \cos{t}\\ y' - \frac{2}{t}y& = t^2 \cos{t}\\ \\ p(t) &= - \frac{2}{t}\\ g(t) &= t^2 \cos{t}\\ \\ \mu (t) &= e^{\int p(t) \; dt}\\ &= e^{\int -\frac{2}{t} \; dt}\\ &= t^{-2}\\ \\ y&= \frac{1}{\mu} \left ( \int \mu.g \;dt + C \right )\\ &= t^2 \left ( \int \cos{t} \;dt + C \right )\\ &= t^2 \left ( \sin{t} + C \right )\\ \end{align*}$

Exact ODEs
Suppose we have a first order linear equation:
$\text{ I) }M(x,y) + N(x,y)\;y' = 0$
where the functions M, N, M_y and N_x are all continuous in the rectangular region R: (x,y) ∈ (α, β) x (γ, δ).

Then, I) is an exact ODE if:
$\text{ II) }M_y (x,y) = N_x(x,y), \forall (x,y) \in R$

That is, there exists a function $\Psi$ satisfying the conditions:
$\text{III) }\Psi_x(x,y) = M(x,y), \Psi_y(x,y) = N(x,y)$
iff M and N satisfy II) $\Psi_y(x,y) = c$ is the implicit solution, and c is a constant.

  Example 6.5 - Exact First ODEs I

Show that the given ODE is an exact ODE, and find the general solution of the given ODE:
$\begin{align*} &\text{A. } y' = -\frac{x+4y}{4x-y}\\ &\text{B. } (3y^3 e^{3xy}-1) + (2ye^{3xy}+3xy^2e^{3xy})y'=0\\ \end{align*}$

Solution:

1. Step 1: Determining M and N:
   $\begin{align*} y' &= -\frac{x+4y}{4x-y}\\ (x+4y)+(4x-y)y' &= 0\\ \\ M&=x+4y\\ M_y&=4\\ N&=4x-y\\ N_x&=4 \end{align*}$
   
   This ODE is indeed an exact ODE, since M_y=N_x.
   
   Step 2: Integrate with respect to y.
   $\begin{align*} \Psi_y (x,y) &= 4x-y \\ \\ \int \Psi_y (x,y) \; dy &= \int (4x-y) \; dy\\ &= 4xy - \frac{1}{2}y^2+C(x) \end{align*}$
   
   Step 3: Using derivative of Ψ with respect to x and M, we can find C(x).
   $\begin{align*} \Psi (x,y) &= 4xy - \frac{1}{2}y^2+C(x)\\ \Psi_x (x,y) &=4y + C'(x)\\ x+4y &= 4y+C'(x)\\ C'(x)&=x\\ \\ C(x) &= \frac{1}{2}x^2+k \end{align*}$
   
   Step 4: Substitute C(x).
   $\Psi (x,y) = 4xy - \frac{1}{2}y^2+\frac{1}{2}x^2+k$
   
   Our final solution is implicitly given by:
   $x^2+8xy-y^2=c$


2. Step 1: Determining M and N:
   $\begin{align*} M&=(3y^3 e^{3xy}-1)\\ M_y&=9y^2e^{3xy}+9xy^2e^{3xy}\\ N&=(2ye^{3xy}+3xy^2e^{3xy})\\ N_x&=6y^2e^{3xy}+{3y^2 e^{3xy}+9xy^3e^{3xy}}\\ &=9y^2e^{3xy}+9xy^2e^{3xy}\\ \end{align*}$
   
   This ODE is indeed an exact ODE, since M_y=N_x.
   
   Step 2: Integrate with respect to x.
   $\begin{align*} \Psi_x (x,y) &= (3y^3 e^{3xy}-1) \\ \\ \int \Psi_x (x,y) \; dx &= \int (3y^3 e^{3xy}-1) \; dx\\ &= y^2 e^{3xy} - x + C(y) \end{align*}$
   
   Step 3: Using derivative of Ψ with respect to y and N, we can find C(y).
   $\begin{align*} \Psi (x,y) &= 3y^3 e^{3xy} - 1 + C(y) \\ \Psi_y&=3y^3 e^{3xy} - 1 + C'(y)\\ C'(y)&=0\\ C(y)&=k \end{align*}$
   
   Step 4: Substitute C(x).
   $\Psi (x,y) = y^2 e^{3xy} - x + k$
   
   Our final solution is implicitly given by:
   $y^2 e^{3xy} - x =c$

  Example 6.6 - Exact First ODEs II

Determine the value of constants a and b so that the following ODE is an exact ODE:
$(y\cos{x}+axe^y)+(b\sin{x}+x^2e^y-1)y'=0$

Solution:
We know that, in an exact ODE, that:
$\begin{align*} M_y &= N_x\\ \cos{x} + axe^y &= b\cos{x} + 2xe^y\\ \\ \cos{x} &= b\cos{x}\\ b&=1\\ \\ axe^y &= 2xe^y\\ a&=2 \end{align*}$

So, the appropriate value of a and b respectively are 2 and 1.