Solve The Given Integral Equation Or Integro-Differential Equation For Latest 2023

Differential equations
Navier–Stokes differential equations used to simulate airflow around an obstruction
Scope
Fields Natural sciences Engineering Astronomy Physics Chemistry Biology Geology Applied mathematics Continuum mechanics Chaos theory Dynamical systems Social sciences Economics Population dynamics List of named differential equations
Classification
Types Ordinary Partial Differential-algebraic Integro-differential Fractional Linear Non-linear By variable type Dependent and independent variables Autonomous Coupled / Decoupled Exact Homogeneous / Nonhomogeneous Features Order Operator Notation
Relation to processes Difference (discrete analogue) Stochastic Stochastic partial Delay
Solution
Existence and uniqueness Picard–Lindelöf theorem Peano existence theorem Carathéodory’s existence theorem Cauchy–Kowalevski theorem
General topics Initial conditions Boundary values Dirichlet Neumann Robin Cauchy problem Wronskian Phase portrait Lyapunov / Asymptotic / Exponential stability Rate of convergence Series / Integral solutions Numerical integration Dirac delta function
Solution methods Inspection Method of characteristics Euler Exponential response formula Finite difference (Crank–Nicolson) Finite element Infinite element Finite volume Galerkin Petrov–Galerkin Green’s function Integrating factor Integral transforms Perturbation theory Runge–Kutta Separation of variables Undetermined coefficients Variation of parameters
People
List Isaac Newton Gottfried Leibniz Jacob Bernoulli Leonhard Euler Józef Maria Hoene-Wroński Joseph Fourier Augustin-Louis Cauchy George Green Carl David Tolmé Runge Martin Kutta Rudolf Lipschitz Ernst Lindelöf Émile Picard Phyllis Nicolson John Crank
v t e

In mathematics, an integro-differential equation is an equation that involves both integrals and derivatives of a function.

General first order linear equations

The general first-order, linear (only with respect to the term involving derivative) integro-differential equation is of the form

d

d
x



u
(
x
)
+

∫


x

0




x


f
(
t
,
u
(
t
)
)

d
t
=
g
(
x
,
u
(
x
)
)
,

u
(

x

0


)
=

u

0


,


x

0


≥
0.


{\displaystyle \frac ddxu(x)+\int _x_0^xf(t,u

differential equations, obtaining a closed-form solution can often be difficult. In the relatively few cases where a solution can be found, it is often by some kind of integral transform, where the problem is first transformed into an algebraic setting. In such situations, the solution of the problem may be derived by applying the inverse transform to the solution of this algebraic equation.

Example

Consider the following second-order problem,

u
′

(
x
)
+
2
u
(
x
)
+
5

∫

0


x


u
(
t
)

d
t
=
θ
(
x
)


with


u
(
0
)
=
0
,


{\displaystyle u'(x)+2u(x)+5\int _0^xu

θ
(
x
)
=

{




1
,

x
≥
0




0
,

x
<
0








{\displaystyle \theta (x)=\left\\beginarrayll1,\qquad x\geq 0\\0,\qquad x<0\endarray\right.

is the Heaviside step function. The Laplace transform is defined by,

$$


U
(
s
)
=


L





u
(
x
)



=

∫

0


∞



e

−
s
x


u
(
x
)

d
x
.


\displaystyle U(s)=\mathcal L\left\u(x)\right\=\int _0^\infty e^-sxu(x)\,dx.

Upon taking term-by-term Laplace transforms, and utilising the rules for derivatives and integrals, the integro-differential equation is converted into the following algebraic equation,

$$


s
U
(
s
)
−
u
(
0
)
+
2
U
(
s
)
+


5
s


U
(
s
)
=


1
s


.


\displaystyle sU(s)-u(0)+2U(s)+\frac 5sU(s)=\frac 1s.

Thus,

$$


U
(
s
)
=


1


s

2


+
2
s
+
5





\displaystyle U(s)=\frac 1s^2+2s+5

.

Inverting the Laplace transform using contour integral methods then gives

$$


u
(
x
)
=


1
2



e

−
x


sin
⁡
(
2
x
)
θ
(
x
)


\displaystyle u(x)=\frac 12e^-x\sin(2x)\theta (x)

.

Alternatively, one can complete the square and use a table of Laplace transforms (“exponentially decaying sine wave”) or recall from memory to proceed:

$$


U
(
s
)
=


1


s

2


+
2
s
+
5



=


1
2




2

(
s
+
1

)

2


+
4



⇒
u
(
x
)
=



L



−
1





U
(
s
)



=


1
2



e

−
x


sin
⁡
(
2
x
)
θ
(
x
)


\displaystyle U(s)=\frac 1s^2+2s+5=\frac 12\frac 2(s+1)^2+4\Rightarrow u(x)=\mathcal L^-1\left\U(s)\right\=\frac 12e^-x\sin(2x)\theta (x)

.

Applications

Integro-differential equations model many situations from science and engineering, such as in circuit analysis. By Kirchhoff’s second law, the net voltage drop across a closed loop equals the voltage impressed

E
(
t
)


{\displaystyle E

L


d

d
t



I
(
t
)
+
R
I
(
t
)
+


1
C



∫

0


t


I
(
τ
)
d
τ
=
E
(
t
)
,


{\displaystyle L\frac ddtI