Example: 1  
Solve the DE 
Solution: 1  
Step1
The I.F. is

Step2
Multiplying by the I.F. on both sides, we have

Step3
Integrating both sides gives

Example: 2  
Solve the DE 
Solution: 2  
Step1
We have,
Note that since the RHS contains the term this DE is not in the standard linear DE form. However, a little artifice can enable us to reduce this to the standard form.
Divide both sides of the equation by :

Step2
Substitute :
Using in , we have

Step3
This is now in the standard firstorder linear DE form. The I.F. is
Thus, the solutions to is

Step4
Performing the integration on the RHS by the substitution and then using integration by parts, we obtain
This is the required general solution to the DE.

This example also tells us how to solve a DE of the general form:
We divide by on both sides :
and then substitute and proceed as described in the solution above.
DEs that take the form in are known as Bernoulli’s DEs.
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In the last section, we discussed how the multiplication of the on both sides of the linear DE
renders this into an exact DE. We now consider the general case of exact DEs. In particular, we want to see what condition must be satisfied in order that the DE
In order for this DE to be exact, it’s LHS must be expressible as the complete differential of some function , i.e.
Now since the function is a function of both and , its total differential is a sum of partical differentials with respect to and , i.e.,
Comparing and , we have
This gives
For continuous ,
and thus for the DE in to be exact, we see that the necessary (and in fact sufficient) condition is
If this condition is satisfied, the DE in reduces to
which upon integration leads to the required solution:
As an example, consider the DE
We have,
Since , the DE is exact and hence we can find a function such that the DE is expressible as . Let us try to explicitly find this function.
From , we have
Integrating with respect to , while treating as a constant, we have
The function acts as the arbitrary constant of integration, since is constant for this integration process.
From again we have
Evaluating from , we have
Finally, substituting into , we have
Thus, the solution to the DE in is
You are urged to verify that is indeed the required solution by differentiating and observing that is obtained.
Example: 3  
Solve the DE 
Solution: 3  
Step1
First of all, notice that this DE is homogeneous :
The substitution leads to

Step2
The substitution leads to

Step3
Integrating both sides gives

Solution: 3 (Alternate) 
Step1
We now solve this DE again using the exact differential approach since by observation this DE satisfies the required criterion for it to be exact. We have
Integrating the first relation, we have

Step2
Differentiating this w.r.t. and comparing with the expression for above, we have
Thus, from and ,

Step3
The solution to the exact DE is
which is the same as the one obtained earlier. Thus, the exact differential approach might lead to the solution faster than the other approaches we’ve discussed earlier.

Sometimes, the fact that the DE is exact is evident merely be inspection. We list down such exact differentials (verify the truth of these relations):