Module 4
First Order Ordinary Differential Equations
A differential equation which is obtained from its primitive by differentiation only and without any operation of elimination and reduction is called an exact differential equation.
Condition: Mdx + Ndy=0 to be exact 
Rule 1: Integrate Mdx w.r.t x treating y constant.
(treating y constant) + 
= c
Rule 2 : Integrate N dy wrt y treating constant .
(treating x constant) + 
= c
Example 1:
Solve (
)
Solution
We have 
and 
and 
Hence, the given equation is exact.
Now, 
And 
The Solution is 
Sometimes a given equation is not exact but is rendered exact if it is multiplied by a suitable factor. If it is called as IF (Integrating Factor).
Equation reducible to exact by IF
Example 1:
Solve 
Solution:
We have
and 
and 
=
I.F. 
Multiplying by 
we get 
, which is exact.
And 
Example 2:
Solve 
Solution:
We have 
and 
I.F 
Multiplying the given equation by
, we get 
which is exact
The solution is 
Example 3:
Solve 
Solution:
We have, 
and 
And 
I.F 
Multiplying the given equation by I.F, we get
which is exact
The solution is 
Example 4:
Solve 
Solution:
The equation can be written as
The equation is of the above form and
Dividing the equation by 
, we get
Now, 
And 
The solution is 
Example 5:
Solve 
Solution:
The equation is homogeneous and
Hence, 
is an integrating factor.
Dividing by 
, we get 
Now this is an exact differential equation.
(By partial fraction treating y constant. Let 
)
(Terms in N free from x) 
The solution is 
Example 6:
Solve 
Solution:
The equation is homogeneous and
Hence, 
is an integrating factor.
Dividing by 
we get 
By partial fraction as in (A) of Ex.3 above)
(Terms in N free from x) 
The solution is 
A differential equation is said to be linear if the dependent variable and its derivative appear only in the 1st degree. The form of linear equation of 1st order is
Steps to solve linear equation
Problem :
Solve 
Solution:
Step -1:
Get the equation to the format
Divide the given equation by 
, We get
Now, 
and 
Step – 2:
To solve the integrating partial fraction is followed,
Put 
Put 
Step – 3:
I.F 
The solution is 
(i) Equation reducible to linear form:
The solution is 
Problem:
Solve 
Solution:
Equation can be written as
and 
Solution is
To solve put 
(ii) Equation solvable for y (By substitution)
Step 1: Put 
Then 
Step 2:
Equation reduces to 
which is linear.
Solution is 
(iii) Equation solvable for x (By substitution)
Solution is 
Example :
Solve 
Solution:
The equation can be written as 
, which is linear.
Now, 
The solution is 
The solution is 
Example :
Solve 
Solution:
We have 
This is linear of the form 
I.F
The solution is 
Put 
The solution is 
Example :
Solve 
Solution:
The equation can be written as
The solution is
The solution is 
Example :
Solve 
Solution:
This equation can be written as
It is of the form 
Now, 
[Put 
]
The solution is 
[Put
]
Example :
Solve 
Solution:
The equation can be written as
Dividing by 
, we get 
Now put 
This is linear of the form 
I.F. 
The solution is
Put 
, 
.
The solution is 
Example:
Solve 
Solution:
The equation can be written as 
Put 
This is a linear differential equation
I.F
The solution is 
[By parts]
.
Step 1: Divide the above equation by
,
Step 2:
Step 3:
Put 
Step 4:
Step 2 becomes
Example:
Solve 
or 
Solution:
The equation can be written as 
Putting 
This is a linear equation
The solution is
Example:
Solve 
Solution:
We have 
Putting 
and 
we get 
This is a linear differential equation
The solution is 
The solution is 
Example:
Solve 
Solution:
Putting 
,
The equation then becomes
This is a Bernoulli’s equation. Dividing by 
throughout, we get
Putting, 
, 
, we get
This is a linear equation.
I.F. 
The solution is 
For integration, we put 
, then 
Hence the solution is 
Putting
, we get the solution as 
In mathematics, a differential equation of the form y = x (dy/dx) + f(dy/dx) where f(dy/dx) is a function of dy/dx only.
The Clairaut equation has the form:
y=xy′+ψ(y′),
where ψ(y′) is a nonlinear differentiable function. The Clairaut equation is a particular case of the Lagrange equation when φ(y′)=y′. It is solved in the same way by introducing a parameter. The general solution is given by
y=Cx + ψ(C),
where C is an arbitrary constant.
Similarly to the Lagrange equation, the Clairaut equation may have a singular solution that is expressed parametrically in the form:
{x=−ψ′(p)y=xp + ψ(p),
where p is a parameter.
Example 1.
Find the general and singular solutions of the differential equation y=xy′+(y′)2.
Solution.
This is a Clairaut equation. By setting y′=p, we write it in the form
y=xp+p2.
Differentiating in x,, we have
dy=xdp+pdx+2pdp.
Replace dy with pdx to obtain:
pdx=xdp+pdx+2pdp,⇒dp(x+2p)=0.
By equating the first factor to zero, we have
dp=0,⇒p=C.
Now we substitute this into the differential equation:
y=Cx+C2.
Thus, we obtain the general solution of the Clairaut equation, which is an one-parameter family of straight lines.
By equating the second term to zero we find that
x+2p=0,⇒x=−2p.
This gives us the singular solution of the differential equation in parametric form:
{x=−2py=xp+p2.
By eliminating p from this system, we get the equation of the integral curve:
p=−x2,⇒y=x(−x2)+(−x2)2=−x22+x24=−x24.
