Differential Equations
Differential equation is a type of equation which contains derivatives in it. The derivative may de
partial deerivative or a ordinary derivative.The eqution may contain derivative of any order.It the
equation contain [dy/dx] then it is a differential equation of first order or first order differential
equation,if the equation contain (d^2y)/(dx^2)] then it is second order differential equation and
so on for higher order differential equations.
Differential equations has wide range of application ,one of the famous application is in newtons
2nd law,it gives us the equation F=ma ,where a is the acceleration.Here a is rate of change of
velocity with respect to time or second derivative of position.If v is the velocity then a= [(dv)/dt]
or in terms of position a= [(d^2u)/(dt^2)] , where u is the position.
In differential equations there are different methods to find the solution for the given differential
equation. One of the methods to solve the differential equation is known as variable separable
method. Now let us see few examples to solve differential equation using variable separable
method.


Learn More about Solving Equations

---

Solving Differential Equations
Back to Top
Below are the examples on solving differential equations using variable separable method.
Example 1 :-
Solve the following differential equation using variable separable method.
[ dy/dx] = 3x + 7.
Solution :-
Given: [dy/dx] = 3x + 7
dy = (5x + 7) dx
Here, the variables are separated, so let's integrate.
[int] dy = [int] (3x + 7 ) dx
y = 3 [((x^2)/(2))] + 7x + c.
Here, c is the constant of integration.


Read More on Graphing Linear Equations

---

Example 2 :-
Solve the following differential equation using variable separable method.
[ dy/dx] = sinx - 3x.
Solution :-
Given: [ dy/dx] = sinx - 3x
dy = (sinx -3 x) dx
Here, the variables are separated, so let's integrate.
[int] dy = [int] ( sinx - 3x) dx
y = -cosx - 3 [((x^2)/(2))] + c.
Hence, the solution of the differential equation.
Example 3 :-
Solve the following differential equation using variable separable method.
(x +1) [dy/dx ] = x2.

---

Solution :-
Given: (x +1) [dy/dx] = x2
[ dy/dx] = [((x^2)/ (x +1))]
[ dy/dx] = x - 1 + [1/(x +1)]
dy = [x - 1 + [1/(x +1)] ] dx
Here, the variables are separated, so let's integrate.
[int] dy = [int] [ x - 1 + [1/(x+1)] ]dx, y = [ (x^2)/(2)] -x + ln(x +1) + c.
Example 4 :-
Solve the following differential equation using variable separable method.
[dy/dx] + y = 1.
Solution :-
Given: [dy/dx] + y = 1
[ dy/dx] = 1 - y
[(dy)/(1-y) ] = dx


Here, the variables are separated, so let's integrate.

---

[int] [(dy)/(1-y) ] = [int dx]
ln(1-y) = x + c
1 - y = e(x +c)
= ex ec
1 - y = k ex
Therefore, the solution is y = 1 - k ex.

---

Thank You
TutorVista.com

---

Document Outline

• 
• ﾿
• 
• ﾿

---