This 6-page document provides a guide to solving differential equations for electrical circuit analysis. It begins by dividing differential equations into homogeneous and non-homogeneous categories. Homogeneous equations have one solution, while non-homogeneous equations have two solutions: the complementary solution and particular solution. The complete solution is the sum of these. The document then discusses finding the complementary solution through the characteristic equation for 1st and 2nd order differential equations. It also covers finding the particular solution depending on whether the right-hand side is a polynomial, exponential, or combination of exponential and trigonometric terms. An example circuit problem is worked through to demonstrate the process.

1. Page 1 of 6
Guide to solve differential equations for electrical circuit analysis
The differential equations can be divided into two main categories.
1. Homogeneous
2. Non-Homogeneous
Homogeneous equations have only one solution. Non-Homogeneous equations have two
solutions named ‘Complementary solution’ and ‘Particular solution’. The total or the
complete solution is the sum of both solutions. If the differential equation represents an
electrical circuit, the complementary solution gives the transient response and the particular
solution gives the steady state response.
At this level we are considering only 1st order and 2nd order equations only.
To find the complementary solution which will be the only solution for Homogeneous
equations, it is required to find the ‘Characteristic equation’. (‘Auxiliary equation’ is another
name). Consider the following example.
If f(t) is a function of t, the 1st order differential equation will be like,
𝐾1
𝑑𝑓(𝑡)
𝑑𝑡
+ 𝐾2 𝑓( 𝑡) = 𝑅. 𝐻. 𝑆
The characteristic equation can be obtained by replacing the differential operator by a
variable. In this example we are using𝛼.
𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ; 𝐾1 𝜎 + 𝐾2 = 0
The root of the characteristic equation is
𝜎 = −
𝐾2
𝐾1
Therefore the complementary solution is,
𝑓𝑐( 𝑡) = 𝐴𝑒 𝜎𝑡
A is a constant.
Now we will find the characteristic equation of a 2nd order differential equation.
𝐾1
𝑑2
𝑓(𝑡)
𝑑𝑡2
+ 𝐾2
𝑑𝑓(𝑡)
𝑑𝑡
+ 𝐾3 = 𝑅. 𝐻. 𝑆
𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ; 𝐾1 𝜎2
+ 𝐾2 𝜎 + 𝐾3 = 0
The second order equations have a quadratic equation as its characteristic equation, therefore
have two roots. It can be found by the formulae.
𝜎 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
‘a’, ‘b’ and ‘c’ represent the cofactors of σ2, σ1 and σ0.

2. Page 2 of 6
The roots can have three different types which will form different types of complementary
solutions. We will discuss all three types.
Case I √𝒃 𝟐 − 𝟒𝒂𝒄 > 𝟎
Roots are real and distinct 𝜎1 and 𝜎2
Therefore
𝒇 𝒄( 𝒕) = 𝑨𝒆 𝝈 𝟏 𝒕
+ 𝑩𝒆 𝝈 𝟐 𝒕
The response is said to be ‘Over damped’
Case II √𝒃 𝟐 − 𝟒𝒂𝒄 = 𝟎
Roots are real and equal 𝜎
Therefore
𝒇 𝒄( 𝒕) = 𝑨𝒆 𝝈𝒕
+ 𝑩𝒕𝒆 𝝈𝒕
The response is said to be ‘Critically damped’
Case III √𝒃 𝟐 − 𝟒𝒂𝒄 < 𝟎
Roots set a complex conjugate pair (𝜎 + 𝑗𝜔) 𝑎𝑛𝑑 (𝜎 − 𝑗𝜔)
Therefore
𝒇 𝒄( 𝒕) = 𝒆 𝝈𝒕
[𝑨𝐜𝐨𝐬 𝝎𝒕 + 𝑩 𝐬𝐢𝐧 𝝎𝒕]
The response is said to be ‘Under damped’
Now we will try to find the particular solution 𝑓𝑝(𝑡) which will be needed only for Non-
Homogeneous equations.
Consider the following examples
𝐾1
𝑑𝑓(𝑡)
𝑑𝑡
+ 𝐾2 𝑓( 𝑡) = 𝑃𝑛(𝑡)
𝐾1
𝑑2
𝑓(𝑡)
𝑑𝑡2
+ 𝐾2
𝑑𝑓(𝑡)
𝑑𝑡
+ 𝐾3 = 𝑃𝑛(𝑡)
R.H.S of the equation is no longer equal to zero. The type of the particular solution depends
on the type of R.H.S. We can represent any type of function on R.H.S. using the following
format.
𝑅. 𝐻. 𝑆. = 𝑒 𝛼𝑡
𝑃𝑛(𝑡)
The function is a product of an exponential and an nth degree polynomial.
Here also we have to consider three different conditions.

3. Page 3 of 6
Case I α is not a root of the characteristic equation.
𝑓𝑝( 𝑡) = 𝑒 𝛼𝑡
𝑄 𝑛 (𝑡)
Same degree polynomial having different cofactors with same exponential.
Case II α is a single (simple) root of the characteristic equation.
𝑓𝑝( 𝑡) = 𝑒 𝛼𝑡
𝑄 𝑛( 𝑡). 𝑡
Polynomial of degree n+1 having different cofactors with same exponential.
Case III α is a double root of the characteristic equation.
𝑓𝑝( 𝑡) = 𝑒 𝛼𝑡
𝑄 𝑛( 𝑡). 𝑡2
Polynomial of degree n+2 having different cofactors with same exponential.
We will now consider another form of R.H.S.
𝑅. 𝐻. 𝑆. = 𝑒 𝛼𝑡
𝑃( 𝑡) cos 𝛽𝑡 + 𝑒 𝛼𝑡
𝑄( 𝑡) sin 𝛽𝑡 …… … …… … (1)
Again there are two types of solutions.
Case I (α + iβ) is not a root of the characteristic equation.
𝑓𝑝( 𝑡) = 𝑒 𝛼𝑡
𝑈 𝑛( 𝑡) cos 𝛽𝑡 + 𝑒 𝛼𝑡
𝑉𝑛( 𝑡)sin 𝛽𝑡
The two polynomials U(t) and V(t) are of the order equal to the highest order of the two
polynomials P(t) and Q(t).
Case II (α + iβ) is a root of the characteristic equation.
𝑓𝑝( 𝑡) = 𝑒 𝛼𝑡
𝑈 𝑛 ( 𝑡). 𝑡 cos 𝛽𝑡 + 𝑒 𝛼𝑡
𝑉𝑛( 𝑡). 𝑡 sin 𝛽𝑡
The two polynomials U(t) and V(t) are of the order equal to the highest order of the two
polynomials P(t) and Q(t).
The sine and cosine terms of the solution stands even if one of the two components (sin or
cosine terms) of the equation (1) is not available in R.H.S.
Up to this point we have discussed how to obtain the complete solution of both
Homogeneous and Non-Homogeneous 1st and 2nd order differential equations.
Let us work out an example
The following example will show the process of forming a differential equation and how the
solution will look like.
VC(t)
R
CV(t) i(t)
Figure 1

4. Page 4 of 6
1. Form the differential equation for the circuit.
Apply KVL to the circuit,
𝑉𝑅 ( 𝑡) + 𝑉𝐶 ( 𝑡) = 𝑉( 𝑡)
𝑅 𝑖( 𝑡) +
1
𝐶
∫ 𝑖( 𝑡) 𝑑𝑡 = 𝑉( 𝑡).. . . . . . . . . . ; {
𝑑𝑞
𝑑𝑡
= 𝐶
𝑑𝑉𝐶
𝑑𝑡
→ 𝑖( 𝑡) = 𝐶
𝑑𝑉𝐶
𝑑𝑡
}
Differentiating above equation w.r.t ‘t’,
𝑅
𝑑𝑖(𝑡)
𝑑𝑡
+
1
𝐶
𝑖( 𝑡) =
𝑑𝑉(𝑡)
𝑑𝑡
Thus, a first order differential equation with constant coefficients describes the
behaviour of the current in this RC circuit.
2. Find the complete response of the circuit by solving the differential equation.
If the applied voltage is,
a) 𝑉( 𝑡) = 𝐸 ; a constant (e.g.: DC source)
𝑅
𝑑𝑖(𝑡)
𝑑𝑡
+
1
𝐶
𝑖( 𝑡) = 0
Since the R.H.S. is equal to zero, the equation is a 1st order Homogeneous equation.
As we discussed the only solution available to this equation is the complementary
solution. It is of the form,
𝑖( 𝑡) = 𝑖 𝑐(𝑡) = 𝐾𝑒 𝜎𝑡
The characteristic equation is,
𝑅𝜎 +
1
𝐶
= 0
∴ 𝜎 = −
1
𝑅𝐶
∴ 𝑖( 𝑡) = 𝐾𝑒
−𝑡
𝑅𝐶⁄
b) 𝑉( 𝑡) = 𝐸 cos 𝜔𝑡 ; (a sinusoidal signal input)
𝑅
𝑑𝑖(𝑡)
𝑑𝑡
+
1
𝐶
𝑖( 𝑡) = −𝐸𝜔 sin 𝜔𝑡
Since the R.H.S is not equal to zero, the equation is Non-Homogeneous.
The characteristic equation is,
𝑅𝜎 +
1
𝐶
= 0 ⟹ ∴ 𝜎 = −
1
𝑅𝐶

5. Page 5 of 6
You can see that the characteristic equation does not change with the R.H.S. because
we equate it to zero to form the characteristic equation.
Therefore the complementary solution stands same for any kind of input to the circuit.
∴ 𝑖 𝐶( 𝑡) = 𝐾𝑒
−𝑡
𝑅𝐶⁄
In order to find the particular solution we have to adjust the R.H.S to fit in to the
general format.
𝑅. 𝐻. 𝑆. = −𝐸𝜔 sin 𝜔𝑡 = 𝑒0𝑡
𝑃0(𝑡)sin 𝜔𝑡
You can see the factor of exponential is zero. It is not a root of the characteristic
equation. The polynomial is a zero degree one which is a constant. Therefore the
solution will be of the form,
𝑖 𝑃( 𝑡) = 𝐴 cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡
The complete solution is,
𝑖( 𝑡) = 𝑖 𝑐( 𝑡)+ 𝑖 𝑝( 𝑡)
𝑖( 𝑡) = 𝐾𝑒
−𝑡
𝑅𝐶⁄
+ [𝐴cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡]
The above example gives an important fact about the solutions of a differential equation. It is
clear that the complementary solution does not depend on the sources or inputs given to the
circuit. It depends only on the elements of the circuit. This means it describes the natural
behaviour of the elements. Therefore the complementary solution represents the Natural
Response of the circuit. Since it is found by equating the R.H.S. to zero, the natural response
is also called Zero Input Response.
The particular solution always depends on the type of input given to the circuit. Therefore it
is called the Forced Response. This means the output is forced to give a certain type by
applying an external source.
As we discussed earlier the complete response is the sum of both solutions. All circuits
respond in its natural behaviour at every excitation. Then it comes to a steady state which will
be governed by the external source.
Now we know how to obtain the solution for a differential equation. After obtaining the
solution it is required to find the values of constants we have applied in the solution.
Following things must be known in order to find the constants.
Initial conditions – values of,
𝑖( 𝑡) ,
𝑑𝑖(𝑡)
𝑑𝑡
and
𝑑2
𝑖(𝑡)
𝑑𝑡2
at 𝑡 = 0+
Example – The initial voltage of a capacitor must be known to find the initial current 𝑖(𝑡)
Important – At excitations (Change of states),
Capacitors block the direct application of voltages across it but allow flowing currents.
Inductors block the flow of current but allow applying voltages across it.

6. Page 6 of 6
References
“Wolfram Alpha Differential equation solver”
http://www.wolframalpha.com/examples/DifferentialEquations.html
Use this online tool to solve differential equations. You can verify your answers by using this
tool.
2015, Guide to solve differential equations for electrical circuit analysis by I.A. Premaratne is
licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.