The Unknown Function

Given:
$$\int_0^x f(t)dt + xf(x) = x^2$$

Solution:

Let $F(x) = \int_0^x f(t)dt$.

Then, the given equation can be written as
$$F(x) + xf(x) = x^2$$

Differentiating both sides with respect to $x$ yields
$$f(x) + F'(x) + xf'(x) = 2x$$

Substituting $F'(x)$ with $f(x)$ gives
$f(x) + xf'(x) = 2x$$Rearranging the equation yields$$f'(x) + \frac{2}{x}f(x) = 2$$This is a first-order linear differential equation with the general solution$$f(x) = c_1e^{\frac{2}{x}} + \frac{2x}{e^{\frac{2}{x}}}$$where$c_1\$ is an arbitrary constant.

Table of Contents

## Integro-differential equation

Equation involving both integrals and derivatives of a function
Differential equations

Navier–Stokes differential equations used to simulate airflow around an obstruction
Scope
Classification
Types
 By variable type Autonomous Coupled / Decoupled Exact Homogeneous / Nonhomogeneous
Relation to processes
Solution
Existence and uniqueness
General topics
People

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

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

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,

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

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,

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

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,

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

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,

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

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

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

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:

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

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

${\displaystyle E

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

I
(
t
)

{\displaystyle I

R

\displaystyle R

is the resistance,

$\displaystyle L$

L

\displaystyle L

the inductance, and

$\displaystyle C$

C

\displaystyle C

the capacitance.

The activity of interacting inhibitory and excitatory neurons can be described by a system of integro-differential equations, see for example the Wilson-Cowan model.

### Epidemiology

Integro-differential equations have found applications in epidemiology, the mathematical modeling of epidemics, particularly when the models contain age-structure or describe spatial epidemics.

## Video about Solve The Given Integral Equation Or Integro-Differential Equation For

L-03 || Integro – Differential Equations || Integral Equation || Integral Equation for M.Sc

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