{"id":5733,"date":"2021-09-29T18:47:37","date_gmt":"2021-09-29T13:17:37","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5733"},"modified":"2021-11-21T20:06:39","modified_gmt":"2021-11-21T14:36:39","slug":"solution-of-linear-differential-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solution-of-linear-differential-equation\/","title":{"rendered":"Solution of Linear Differential Equation"},"content":{"rendered":"
Here you will learn how to find solution of linear differential equation of first order first degree with examples.<\/p>\n
Let’s begin –<\/p>\n
A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree.<\/p>\n
\nThe general form of a linear differential equation is <\/p>\n
\\(dy\\over dx\\) + Py = Q<\/p>\n
where P and Q are functions of x (or constants)<\/p>\n<\/blockquote>\n
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.<\/p>\n
Algorithm :<\/strong><\/p>\n
\n1). Write the differential equation in the form \\(dy\\over dx\\) + Py = Q and obtain P and Q<\/p>\n
2). find the integrating factor (I. f.) given by I.f = \\(e^{\\int Pdx}\\)<\/p>\n
3). Multiply both sides of equation in step 1 by I.f.<\/p>\n
4). Integrate both sides of the equation obtained in step 3 with respect to x to obtain <\/p>\n
y(I.f) = \\(\\int\\) Q(I.f) dx + C, which gives the required solution.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Solve the differential equation : \\(dy\\over dx\\) – \\(y\\over x\\) = \\(2x^2\\)<\/p>\n
Solution<\/span><\/strong> : We are given that,<\/p>\n
\\(dy\\over dx\\) – \\(y\\over x\\) = \\(2x^2\\)<\/p>\n
Clearly it is a differential equation of the form<\/p>\n
\\(dy\\over dx\\) + Py = Q , where P = \\(-1\\over x\\) and Q = \\(2x^2\\)<\/p>\n
Now, I.f = \\(e^{\\int Pdx}\\) = \\(e^{\\int (-1\/x)dx}\\) = \\(e^{-log x}\\) = \\(1\\over x\\)<\/p>\n
By algorithm, the solution is<\/p>\n
y\\(1\\over x\\) = \\(\\int\\) 2x dx + C<\/p>\n
\\(\\implies\\) \\(y\\over x\\) = \\(x^2\\) + C<\/p>\n
\\(\\implies\\) y = \\(x^3\\) + Cx, which is the required solution.<\/p>\n
(2) Linear Differential Equation of the form \\(dx\\over dy\\) + Rx = S<\/strong><\/h4>\n
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<\/p>\n
Note that here y is independent variable and x is a dependent variable.<\/p>\n
Algorithm :<\/strong><\/p>\n
\n1). Write the differential equation in the form \\(dx\\over dy\\) + Rx = S and obtain R and S<\/p>\n
2). find the integrating factor (I. f.) given by I.f = \\(e^{\\int Rdy}\\)<\/p>\n
3). Multiply both sides of equation in step 1 by I.f.<\/p>\n
4). Integrate both sides of the equation obtained in step 3 with respect to x to obtain <\/p>\n
x(I.f) = \\(\\int\\) S(I.f) dy + C, which gives the required solution.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Solve the differential equation : ydx + \\(x – y^3\\) dy = 0<\/p>\n
Solution<\/span><\/strong> : We are given that,<\/p>\n
ydx + \\(x – y^3\\) dy = 0<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) + \\(x\\over y\\) = \\(y^2\\)<\/p>\n
Clearly it is a differential equation of the form<\/p>\n
\\(dx\\over dy\\) + Ry = S , where R = \\(1\\over y\\) and S = \\(y^2\\)<\/p>\n
Now, I.f = \\(e^{\\int Rdy}\\) = \\(e^{\\int (1\/y)dy}\\) = \\(e^{log y}\\) = y<\/p>\n
By algorithm, the solution is<\/p>\n
xy = \\(\\int\\) \\(y^3\\) dy + C<\/p>\n
\\(\\implies\\) xy = \\(y^4\\over 4\\) + C, which is the required solution.<\/p>\n\n\n