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# Higher Order Linear Differential equations

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Differentiation is important and interesting part of mathematics. We use differentiation process for calculating the rate of particular function with respect to particular variable like f(x) is a function and d/dx(f(x) is a rate of f(x) with respect to x. Now we discuss different type of differentiation equations –

For understand the different type of differentiation equation we assume a function y = f(x) and when we calculate differentiation of f(x) with respect to x, it produces first order differentiation equation and these type of first order differentiation contains dy/dx and example of first order differentiation equation are – if y = sin x, then dy/dx = cos x

When we differentiate dy/dx with respect to x, it produces second order differentiation equation and these type of second order equation contains d2y/dx2 form like – if y = sin-1 x, then d2y/dx2 = x/(1-x2)3/2

When we differentiate d2y/dx2 with respect to x, it produces third order differentiation equation and these type of third order differentiation equation contains d3y/dx3 form like – if y = log(sin x), then third order differentiation equation is d3y/dx3 = 2 sin x * cosec3 x

Similarly this process continues to produce n order differentiation equation and these type of n order differentiation equation contains dny/dxn form and these type of equations are called as a higher order Differential Equations. A linear equation which contains these higher order Derivatives are called as a higher order linear differential equations.

For calculating the higher order differential equations, we use following steps –

Step 1: First of all we assume given function as a variable like y = f(x).

Step 2: After assuming that variable, we differentiate that variable with respect to any suitable variable like we differentiate y with respect to x, it produces dy/dx.

Step 3: According to the equation we again differentiate variable with respect to that particular variable like if we differentiate dy/dx with respect to x again, it produces (d2y/dx2), which is a second order differentiation equation.

Step 4: According to situation, we continue this differentiation until we get our required equation.

Step 5: After all differentiate process, we put the values of dy/dx, d2y/dx2,…………dny/dxn in that equation, which we want to prove.

Now we take some examples of higher order differential equation –

Example 1: If y = xx, then find d2y/dx2.

Solution:

Step 1 : First of all we assume y = xx which is given in above question. Therefore –

y = xx

After taking log both side, we get –

log y = x log x

Step 2: Now we differentiate both side with respect to x and we get

d/dx(log y) = d/dx(x log x)

=> (1/y)dy/dx = x * d/dx(log x) + log x * d/dx(x)

=> (1/y)dy/dx = x * 1/x + log x * 1

=> (1/y)dy/dx = 1 + log x

=> dy/dx = y(1 + log x) ……………..equation(1)

Step 3: Now we differentiate equation (1) with respect to x, we get

d2y/dx2 = d/dx(y(1 + log x)) = [y * d/dx(1 + log x) + (1 + log x) * d/dx(y)]

=> d2y/dx2 = y * 1/x + (1 + log x) * (dy/dx)

After putting the value of dy/dx from eq(1)

=> d2y/dx2 = y * 1/x + (1 + log x) * (y(1 + log x))

=> d2y/dx2 = y /x + y * (1 + log x)2

After putting the value of y, we get:

=> d2y/dx2 = xx/x + xx * (1 + log x)2

Now we take common part xx from above equation,

=> d2y/dx2 = xx [ 1/x + (1 + log x)2].

Example 2: If y = x +√x2 + 1m, then shows that (x2 + 1)*d2y/dx2 + x *(dy/dx) – m2y = 0

Solution:

Step 1: First of all we assume y =x +√x2 + 1m which is given in above question.

Step 2: Now we differentiate y with respect to x and we get:

dy/dx = m.( x +√x2 + 1m-1) * d/dx(x +√x2 + 1),

=> dy/dx = m.( x +√x2 + 1m-1) *1 + 2x/x*√(x2 + 1),

=> dy/dx = m* x +√x2 + 1m/√(x2 + 1),

=> dy/dx = my/√(x2 + 1),

=> (√(x2 + 1))*dy/dx = my

Now we squaring both sides

=> (x2 + 1)*(dy/dx)2 = m2 y2 ……………..equation(1)

Step 3: Now we differentiate equation(1) with respect to x, we get:

2 *(dy/dx)* (d2y/dx2)*(1 + x2) + (dy/dx)2 *(2x) = 2m2*y* dy/dx,

=> (x2 + 1)*d2y/dx2 + x *(dy/dx) – m2y = 0,

Hence proved.

Example 3: If x = a cos t + b sin t and y = a sin t – b cos t , then prove that y2 *d2y/dx2 - x *(dy/dx) + y = 0

Solution:

Step 1: First of all we calculate x2 + y2 =>

x2 + y2 => (a cos t + b sin t)2 + (a sin t – b cos t)2

=> a2(cos2 t + sin2 t) + b2(cos2 t + sin2 t)

As we know that cos2 t + sin2 t = 1, then

x2 + y2 = a2 + b2 …………………eq(1)

Step 2: Now we differentiate eq(1) with respect to x and we get

2x + 2y *dy/dx = 0

dy/dx = -x/y …………………eq(2)

Step 3: Now we differentiate equation(2) with respect to x, we get

d2y/dx2 = -(y.1 – x*(dy/dx))/y2

Now we put value of (dy/dx) in above equation

d2y/dx2 = -(y.1 – x*(-x/y))/y2

d2y/dx2 = -(x2 - y2)/y3 ……………….eq(3)

Step 4: Now, we put all values of dy/dx, d2y/dx2 in given equation –

y2 *d2y/dx2 - x *(dy/dx) + y = 0

=> y2 *-(x2 - y2)/y3 - x *(-x/y) + y

=> x2/y – y - x2/y + y

=> 0 = rhs

So, we can say that y2 *d2y/dx2 - x *(dy/dx) + y = 0

Hence proved.

These are some examples of higher order differentiation equation which shows how we calculate higher order differential equations from a simple function and how we can prove higher order differential equations.