Basic Solution Concepts,
& Separable ODEs
Solution of ODE Y to the left….
X to the right…
• General Solution
• Initial Value Solution
Separable ODE
A differential equation (DE)
𝜕𝑦
E.g. 𝑦 ′ = = cos 𝑥
𝜕𝑥
can be solved (in order to find its solution) through
integration
𝑦 = 𝑓 𝑥 = න 𝜕𝑦 = න cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐
i.e. c can be any values e.g.
0, -100, 99, 2.75 etc.
Hence this solution is
known as a family
solution
A functional equation
𝑐
E.g. 𝑦=𝑓 𝑥 =
𝑥
has its own differential equation (DE)
𝜕𝑦 𝑐
i.e. 𝑦′ = = − 2
𝜕𝑥 𝑥
A functional equation can be a solution to its differential
equation (DE)
Differential 𝑐 Functional
𝑥𝑦 ′ = − = −𝑦
Equation 𝑥 equation
This function is a solution to
its differential equation!!!!
The functional equation
E.g. 𝑦 = 𝑓 𝑡 = 𝑐𝑒 0.2𝑡
has its own differential equation (DE)
𝜕𝑦
i.e. 𝑦 ′ = = 0.2(𝑐𝑒 0.2𝑡 )
𝜕𝑡
A functional equation can be a solution to its differential
equation (DE)
Differential 𝜕𝑦 Functional
𝑦′ = = 0.2(𝑐𝑒 0.2𝑡 ) = 0.2𝑦
Equation 𝜕𝑡 equation
This function is a solution to its
differential equation!!!!
The solution y has an arbitrary constant c –
hence solution y is a general solution!!!
Based on example 3, the following ODE:
𝜕𝑦
𝑦′ = = 0.2𝑦
𝜕𝑡
has a general solution of:
𝑦 = 𝑓 𝑡 = 𝑐𝑒 0.2𝑡
What is the solution, given an initial value y of 0.5 at t = 0
(i.e. y(0)=0.5)?
𝑦 = 𝑓 𝑡 = 0 = 𝑐𝑒 0.2 0
= 0.5; 𝑇ℎ𝑢𝑠 𝑦 = 𝑐 = 0.5
Thus, the solution at the initial point is 𝑦 = 𝑓 𝑡 = 0.5𝑒 0.2𝑡
The solution y does NOT contain an
arbitrary constant c at the initial point–
hence solution y is a particular solution at
the initial point!!!
Based on example 3, the following ODE:
𝜕𝑦
𝑦′ = = 0.2𝑦
𝜕𝑡
has a general solution of:
𝑦 = 𝑓 𝑡 = 𝑐𝑒 0.2𝑡
The c value may vary, at initial point:
𝑦=𝑓 𝑡=0 =𝑐
Based on the negative version of example 3, the following
ODE: ′
𝜕𝑦
𝑦 = = −0.2𝑦
𝜕𝑡
has a general solution of:
𝑦 = 𝑓 𝑡 = 𝑐𝑒 −0.2𝑡
The c value may vary, at initial point:
𝑦=𝑓 𝑡=0 =𝑐
Assuming that an ODE consists of 2 different variables x and
y, both variables can be separated as such x is on the right
and y is on the left (vice versa).
𝑔 𝑦 . 𝑦′ = 𝑓 𝑥
By separating x and y, we can integrate both sides
independently in order to derive the general solution:
𝜕𝑦
න𝑔 𝑦 𝜕𝑥 = න 𝑓 𝑥 𝜕𝑥
𝜕𝑥
Please remember to
Thus: always introduce c
𝑥𝜕 𝑥 𝑓 = 𝑦𝜕 𝑦 𝑔 immediately after
integration
The following ODE: 𝑦′ = 1 + 𝑦2
𝑦′
Can be separated into: 1+𝑦 2
=1
The solution through integration: Please remember to
1 always introduce c
න 𝜕𝑦 = න 1 𝜕𝑥 immediately after
1 + 𝑦2
integration –
produces….. arctan 𝑦 = 𝑡𝑎𝑛−1 (𝑦) = 𝑥 + 𝑐 otherwise you will get
different results
Thus: 𝑦 = tan 𝑥 + 𝑐 y = tan (x) + c
Recall: Recall:
𝛿 1 1
−1
𝑡𝑎𝑛 (𝑦) = 𝑡𝑎𝑛−1 𝑦 = = arctan(𝑦)
𝛿𝑥 1 + 𝑦2 tan(𝑦)
𝜕𝑦
The following ODE: 𝑦 ′ = 𝜕𝑥 = 𝑥 + 1 𝑒 −𝑥 𝑦 2
𝑦′
Can be separated into: 𝑦2
= 𝑥 + 1 𝑒 −𝑥
The solution through integration:
1
න 2 𝜕𝑦 = න 𝑥 + 1 𝑒 −𝑥 𝜕𝑥
𝑦
produces…
1
− 1 = න(𝑥 𝑒 −𝑥 + 𝑒 −𝑥 )𝜕𝑥 = න(𝑥 𝑒 −𝑥 )𝜕𝑥 + න 𝑒 −𝑥 𝜕𝑥
𝑦
= −𝑒 −𝑥 + 𝑒 𝑥(−𝑥 ) 𝜕𝑥 Recall:
1
Thus: −
𝑦1
න 𝑢 𝜕𝑣 = 𝑢𝑣 − න 𝑣 𝜕𝑢
1
For: −
𝑦1
= −𝑒 −𝑥 + 𝑒 𝑥(−𝑥 ) 𝜕𝑥 = −𝑒 −𝑥 + 𝑣𝜕 𝑢
Recall:
where: 𝑢 = 𝑥 𝑎𝑛𝑑 𝜕𝑣 = 𝑒 −𝑥 𝜕𝑥
න 𝑢 𝜕𝑣 = 𝑢𝑣 − න 𝑣 𝜕𝑢
Thus: 𝜕𝑢 = 𝑑𝑥 𝑎𝑛𝑑 𝑣 = −𝑒 −𝑥
Substitute back into the original equation:
1
− 1 = −𝑒 −𝑥 + −𝑥𝑒 −𝑥 − න −𝑒 −𝑥 𝑑𝑥 = −𝑒 −𝑥 − 𝑥𝑒 −𝑥 − 𝑒 −𝑥 + 𝑐
𝑦
Hence…….
1
𝑦=
𝑥 + 2 𝑒 −𝑥 − 𝑐
Solved the following ODE as its initial value
(i.e. derive the particular solution):
𝑦 ′ = −2𝑥𝑦 𝑎𝑡 𝑦 0 = 1.8
𝑦′ Recall:
Through separation: 𝑦
= −2𝑥 1 Assume:
න 𝜕𝑦 = ln(𝑦) ℂ = 𝑒𝑐
𝑦
The solution through integration: Through
1 initial value
න 𝜕𝑦 = න −2𝑥 𝜕𝑥 substitution:
𝑦 ℂ=1.8
produces…. ln 𝑦 = −𝑥 2 + 𝑐
2
Thus: 𝑦 = ℂ𝑒 −𝑥
−𝑥 2
The initial value solution: 𝑦= 1.8𝑒
Solve the ODE:
• 𝑦 ′ + 2 sin 2𝜋𝑥 = 0
• 𝑦 ′ = 4𝑒 −𝑥 cos(𝑥)
Verify that y is a solution of the ODE. Solve the IVP.
• 𝑦 ′ + 4𝑦 = 1.4
𝑦 = 𝑐𝑒 −4𝑥 + 0.35 𝑎𝑛𝑑 𝑦 0 = 2
• 𝑦′ = 𝑦 + 𝑒 𝑥
1
𝑦 = 𝑥+𝑐 𝑒𝑥 𝑎𝑛𝑑 𝑦 0 =
2
Find the general solution.
• 𝑦 ′ = 𝑠𝑒𝑐 2 𝑦
• 𝑦𝑦 ′ + 36𝑥 = 0
Find the general solution. Solve
the IVP.
• 𝑥𝑦 ′ + 𝑦 = 0
𝑦 4 =6
−4𝑥
• 𝑦′ =
𝑦
𝑦 2 =3
“Time and tide wait for no
man””
--unknown
