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Differential equations are classified into different types such as Linear differential equations, exact and non-exact differential equations, homogeneous differential equations etc. All such equations can be solved by using various methods.
Certain non-exact differential equations can be directly solved by careful inspection of them. The method used for solving such equations is known as the ‘Method of inspection’.
What is an exact differential equation? –
The equation of type A dp+B dq are called exact differential equations only if they satisfy the condition A/q=B/p.
In this equation p and q are variables and A(p,q) & B(p,q) are two functions.
What is a non-exact differential equation? –
The equation of type A dp+B dq are called non-exact differential equations only if they satisfy the condition A/qB/p.
In this equation p and q are variables and A(p,q) & B(p,q) are two functions.
How to solve a non-exact differential equation?-
Generally, non-exact differential equations are first converted into exact differential equations, and then they are solved. But sometimes, it becomes very difficult to convert a given non-exact differential equation into an exact differential equation. Hence, we can use the ‘Method of inspection’ to solve such non-exact equations in such cases.
Before understanding the ‘Method of inspection,’ we first need to learn the concept of ‘Integrable’.
Integrables –
Integrables are parts of the differential equations whose integration can be done easily.
E.g. d(p2)=p2 , d(pq)=pq etc.
Here is the sign of integration (antiderivative), and d() represents the differential of a function in the argument.
Some important integrables –
- q dp+p dq=d(pq)
- (q dp-p dq)/q2=d(p/q)
- dp+dq=d(p+q)
- d(q/p)=(p dq-q dp)/p2
- d(p2/q)=(2pqdp-p2dq)/q2
- d(q2/p)=(2pqdq-q2dp)/p2
- d(p2/q2)=(2pq2dp-2p2qdq)/q4
- d(q2/p2)=(2qp2dq-2q2pdp)/p4
- d(log(pq))=(p dq+q dp)/pq
- d(log(p/q))=(q dp-p dq)/pq
- d(log(q/p))=(p dq-q dp)/pq
- d(log(p+q))=(dp+dq)/p+q
- d(log(p2+q2))=(p dp+q dq)/(p2+q2)
- d(tan-1(q/p))=(p dq-q dp)/(p2+q2)
- d(tan-1(p/q))=(q dp-p dq)/(p2+q2)
- d(-1/pq)=(p dq-q dp)/p2q2
- d(ep/q)=(qepdq-epdq)/q2
- d(eq/p)=(peqdp-eqdp)/p2
- d(p2+q2)=(p dp+q dq)/p2+q2
- d(pm,qn)=pm-1qn-1(mq dp+np dq)
- d[(1/2)*(log (p+q)/(p-q))]=(p dq-q dp)/(p2–q2)
- d[f(p,q)1-n]/(1-n)=f'(p,q)/[f(p,q)n]
- d(1/q-1/p)=d(1/q)-d(1/p)=dp/p2-dq/q2
Remembering these integrables helps us a lot while solving problems.
Procedure to solve the problem –
Step 1 –
Check whether the given differential equation is exact or not. If it is non-exact, then only proceed with the next step.
Step 2 –
Rearrange the given differential equation so that the integrables are visible.
Step 3 –
Integrate the entire equation to get the final answer.
Let us understand this method with an example.
Solve: p dq-q dp=(p2+q2)dp
Solution: We first check whether the given equation is exact or not.
Given differential equation is
p dq-q dp=(p2+q2)dp …………..(1)
∴ (-q-q2–p2)dp+p dq=0
Comparing this equation with the standard form A dp+B dq=0, we get
A=(-q-q2–p2) & B=p
A/q=(-q-q2–p2)/q=-1-2q
B/p=1
∴ A/qB/p
Therefore, the given differential equation is not exact.
Now, from equation (1) we can write
p dq-q dp=(p2+q2)dp
Dividing both sides by (p2+q2), we get
(p dq-q dp)/(p2+q2)=(p2+q2)dp/(p2+q2)
(p dq-q dp)/(p2+q2)=dp ………..(2)
But as we have seen that
(p dq-q dp)/(p2+q2)=d(tan-1(q/p))
Therefore, equation (2) becomes
∴ d (tan-1(q/p))-dp=0…………….(3)
Equation (3) contains an easy-to-solve integrable. By integrating equation (2), we can get the solution for the given differential equation. Thus, integrating both sides of equation (3) we get
d (tan-1(q/p))-dp=0
∴ d(tan-1(q/p))-dp=0
∴ tan-1(q/p)+A1-p+A2=0
∴ tan-1(q/p)=p+A
Where, A1, A2,A are arbitrary constants.
This is the required solution.
Advantages of inspection method –
- It is an easy method.
- It does not include conversion of non-exact to exact differential equations.
Disadvantages of inspection method –
- One might find it difficult to remember all the integrables.
Conclusion –
In this article, we studied the method of inspection to solve non-exact differential equations. We have also solved one problem with the help of the inspection method.
