Here you will learn how to find solution of linear differential equation of first order first degree with examples.
Let’s begin –
Solution of Linear Differential Equation
(1) Linear Differential Equation of the form \(dy\over dx\) + Py = Q
A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree.
The general form of a linear differential equation is
\(dy\over dx\) + Py = Q
where P and Q are functions of x (or constants)
This type of differential equations are solved when they are multiplied a factor, which is called integrating factor, because by multiplication of this factor the left hand side of the differential equation (i) becomes exact differential of some function.
Algorithm :
1). Write the differential equation in the form \(dy\over dx\) + Py = Q and obtain P and Q
2). find the integrating factor (I. f.) given by I.f = \(e^{\int Pdx}\)
3). Multiply both sides of equation in step 1 by I.f.
4). Integrate both sides of the equation obtained in step 3 with respect to x to obtain
y(I.f) = \(\int\) Q(I.f) dx + C, which gives the required solution.
Example : Solve the differential equation : \(dy\over dx\) – \(y\over x\) = \(2x^2\)
Solution : We are given that,
\(dy\over dx\) – \(y\over x\) = \(2x^2\)
Clearly it is a differential equation of the form
\(dy\over dx\) + Py = Q , where P = \(-1\over x\) and Q = \(2x^2\)
Now, I.f = \(e^{\int Pdx}\) = \(e^{\int (-1/x)dx}\) = \(e^{-log x}\) = \(1\over x\)
By algorithm, the solution is
y\(1\over x\) = \(\int\) 2x dx + C
\(\implies\) \(y\over x\) = \(x^2\) + C
\(\implies\) y = \(x^3\) + Cx, which is the required solution.
(2) Linear Differential Equation of the form \(dx\over dy\) + Rx = S
Sometimes a linear differential equation can be put in the form \(dx\over dy\) + Rx = S where R and S are functions of y or constants
Note that here y is independent variable and x is a dependent variable.
Algorithm :
1). Write the differential equation in the form \(dx\over dy\) + Rx = S and obtain R and S
2). find the integrating factor (I. f.) given by I.f = \(e^{\int Rdy}\)
3). Multiply both sides of equation in step 1 by I.f.
4). Integrate both sides of the equation obtained in step 3 with respect to x to obtain
x(I.f) = \(\int\) S(I.f) dy + C, which gives the required solution.
Example : Solve the differential equation : ydx + \(x – y^3\) dy = 0
Solution : We are given that,
ydx + \(x – y^3\) dy = 0
\(\implies\) \(dy\over dx\) + \(x\over y\) = \(y^2\)
Clearly it is a differential equation of the form
\(dx\over dy\) + Ry = S , where R = \(1\over y\) and S = \(y^2\)
Now, I.f = \(e^{\int Rdy}\) = \(e^{\int (1/y)dy}\) = \(e^{log y}\) = y
By algorithm, the solution is
xy = \(\int\) \(y^3\) dy + C
\(\implies\) xy = \(y^4\over 4\) + C, which is the required solution.
