Mathematics Linear differential equations For CBSE-NCERT 4
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### Topic Covered

color {red} ♦ Linear differential equations
color {red} ♦ Steps involved to solve first order linear differential equation
color {red} ♦ Miscellaneous Examples

### Linear differential equations

\color{green} ✍️ A differential equation of the from

(dy)/(dx) + Py = Q

=> where, P and Q are constants or functions of x only, is known as a first order linear differential equation.

=> Some examples of the first order linear differential equation are

(dy)/(dx) + y = sinx

(dy)/(dx) + (1/x) y = e^x

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

=> Another form of first order linear differential equation is

(dx)/(dy) + P_1x =Q_1

where, P_1 and Q_1 are constants or functions of y only. Some examples of this type of differential equation are

(dx)/(dy) + x = cos y

(dx)/(dy) + (-2x)/y = y^2e^(-y)

\color{green} ✍️ To solve the first order linear differential equation of the type

color {red} {(dy)/(dx) + Py = Q} ......................(1)

=> Multiply both sides of the equation by a function of x say g (x) to get

color {red} { g(x) (dy)/(dx) + P * (g(x) ) y = Q * g(x)}  ................................(2)

=> Choose g (x) in such a way that R.H.S. becomes a derivative of y . g (x).

i.e color {red} { g(x) (dy)/(dx) + P* g(x) y = d/(dx) [ y *g(x) ]}

or g(x) (dy)/(dx) +P * g(x) y = g(x) (dy)/(dx) + y g'(x)

=> P*g(x) = g' (x)

or P = (g'(x))/(g(x))

=> Integrating both sides with respect to x, we get

int Pdx = int (g'(x))/(g(x)) dx

or int P* dx = log (g (x) )

or g(x) = e^(int P dx)

=> On multiplying the equation (1) by g(x) = e^(int P dx) , the L.H.S. becomes the derivative of some function of x and y.

=> This function g(x) = e^(int Pdx) is called Integrating Factor (I.F.) of the given differential equation.

Substituting the value of g (x) in equation (2), we get

e^(int P dx) (dy)/(dx) + P e^(Pdx) y = Q * e^(int P dx)

or  d/(dx) ( y e ^(int P dx)) = Q e^(intPdx)

=> Integrating both sides with respect to x, we get

color{blue}{y*e^(int Pdx) = int (Q * e^(int P dx) )dx}

or color{blue}{y = e^(-int Pdx) * int (Q * e^(int Pdx)) dx +C}

which is the general solution of the differential equation.

### Steps involved to solve first order linear differential equation:

(i) Write the given differential equation in the form (dy)/(dx) + Py = Q where where P, Q are constants or functions of x only.

(ii) Find the Integrating Factor (I.F) = color {red} {e^(int Pdx)}

(iii) Write the solution of the given differential equation as

color {red} {y (I.F) = ∫ (Q × I.F)dx + C}

\color{green} ✍️ In case, the first order linear differential equation is in the form (dx)/(dy) + P_1x =Q_1

=> where, P_1 and Q_1 are constants or functions of y only. Then I.F =e^( ∫ P_1 dy ) and the solution of the differential equation is given by

color {red} {x . (I.F) = ∫ (Q_1 × I.F ) dy + C}
Q 3187156987

Find the general solution of the differential equation (dy)/(dx) - y =cos x
Class 12 Chapter 9 Example 19
Solution:

Given differential equation is of the form

(dy)/(dx) + Py = Q, where P =-1 and Q = cos x

Therefore I.F = e^(int - 1 dx ) = e^(-x)

Multiplying both sides of equation by I.F, we get

e^(-x) (dy)/(dx) - e^(-x) y = e^-x cos x

or (dy)/(dx) (ye^(-x) ) = e^(-x) cos x

On integrating both sides with respect to x, we get

y e^(-x) = int e^(-x) cos x dx +C

Let I = int e^(-x) cos dx

= cos x ((e^-x)/(-1) ) - int (- sin x ) ( -e^(-x) ) dx

= −cos x e^-x − ∫sin x e^-x dx

= - cos x e^(-x) - [ sin x (-e^(-x) ) - int cos x ( -e ^(-x) ) dx ]

= - cos e^(-x) + sin e^(-x) - int cos x e^(-x) dx

or I = - e^(-x) cosx + sin x e^-x - I

or  2I = ( sin x - cos x ) e^(-x)

or  I = (( sin x - cos x ) e^(-x))/2

Substituting the value of I in equation (1), we get

ye^(-x) = (( sin x - cos x )/2 ) e^(-x) +C

or  y = ((sin x - cos x )/2 ) + C e^x

which is the general solution of the given differential equation.
Q 3137167082

Find the general solution of the differential equation x (dy)/(dx) + 2 y = x^2 \ \ \ \ (x != 0)

Class 12 Chapter 9 Example 20
Solution:

The given differential equation is

x (dy)/(dx) + 2 y = x^2 .......................(1)

Dividing both sides of equation (1) by x, we get

(dy)/(dx) + 2/x y =x

(dy)/(dx) + 2/x y =x

which is a linear differential equation of the type (dy)/(dx) + Py =Q, where P = 2/x and Q =x

So I.F = e^( int 2/x dx ) = e^(2 log x^2 ) = x^2 [ text(as ) e^(log f(x) ) = f(x) ]

Therefore, solution of the given equation is given by

y* x^2 = int (x) (x^2 ) dx + C = int x^3 dx +C

or  y = x^2/4 +Cx^-2

which is the general solution of the given differential equation.
Q 3167167085

Find the general solution of the differential equation y dx – (x + 2y^2) dy = 0.

Class 12 Chapter 9 Example 21
Solution:

The given differential equation can be written as

(dx)/(dy) - x/y = 2y

This is a linear differential equation of the type (dx)/(dy) + P_1x = Q_1 , whre P_1 = -1/y and Q_1 = 2y.

Therefore I.F = e^(int - 1/y dy) = e^(-log y) = e^(log (y )^(-1)) = 1/y

Hence, the solution of the given differential equation is

x 1/y = int (2y) ( 1/y ) dy +C

or x/y = int (2 dy ) +C

or x/y = 2y +C

or  x = 2y^2 +Cy

which is a general solution of the given differential equation.
Q 3147667583

Find the particular solution of the differential equation (dy)/(dx) + y cot x = 2 x + x^2 cot x \ \ \ \ (x!=0)

gives that y = 0 when x = pi/2
Class 12 Chapter 9 Example 22
Solution:

The given equation is a linear differential equation of the type (dy)/(dx) + Py = Q

where P = cot x and Q = 2 x +x^2 cot x . Therefore

I.F = e^(int cot x dx ) = e^(log sin x) = sin x

Hence, the solution of the differential equation is given by

y . sin x = ∫(2x + x^2 cot x) sin x dx + C

or y sin x = ∫2x sin x dx + ∫ x^2 cos x dx + C

or  y sin x = sin x((2x^2)/2) - int cos x ((2x^2)/2) dx + int x^2 cos dx +C

or y sin x = x^2 sin x - int x^2 cos dx + int x^2 cos dx +C

or  y sin x = x^2 sin x +C ...............(1)

Substituting y = 0 and x =pi/2  in in equation (1), we get

0 = (pi/2)^2 sin (pi/2 ) +C

or  C = (-pi^2)/4

Substituting the value of C in equation (1), we get

y sin = x^2 sin x - pi^2/4

or  y = x^2 - pi^2/(4 sin x ) \ \ \ \ (sin x != 0)

which is the particular solution of the given differential equation.
Q 3117667589

Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point.
Class 12 Chapter 9 Example 23
Solution:

We know that the slope of the tangent to the curve is (dy)/(dx)

Therefore, (dy)/(dx) = x + xy

or (dy)/(dx) - xy = x ............................(1)

This is a linear differential equation of the type (dy)/(dx) + Py = Q , where P = -x and Q = x

Therefore, I.F = e^(int -x dx ) = e^(-x^2/2)

Hence, the solution of equation is given by

y * e^(-x^2/2) = int (x) ( e^(-x^2/2)) dx +C ..................(2)

Let I = int (x) e^(-x^2/2) dx

Let (-x^2)/2 =t , then -x dx = dt or x dx = -dt

Therefore , I = -int e^t dt = - e^t = - e^(-x^2/2)

Substituting the value of I in equation (2), we get

ye^((-x^2)/2) = - e^(-x^2/2) +C

or y = -1 + C e^(x^2/2) ..................................(3)

Now (3) represents the equation of family of curves. But we are interested in finding a particular member of the family passing through (0, 1). Substituting x = 0 and y = 1 in equation (3) we get

1 = – 1 + C . e^0 or C = 2

Substituting the value of C in equation (3), we get

y = -1 + 2 e^(x^2/2)

which is the equation of the required curve.
Q 3117067880

Verify that the function y = c_1 e^(ax) cos bx + c_2 e^(ax) sin bx, where c_1, c_2 are arbitrary constants is a solution of the differential equation

(d^2y)/(dx^2) - 2 a (dy)/(dx) + ( a^2 + b^2) y = 0
Class 12 Chapter 9 Example 24
Solution:

The given function is

y = e^(ax) [c_1 cosbx + c_2 sinbx] ... (1)

Differentiating both sides of equation (1) with respect to x, we get

(dy)/(dx) = e^(ax) [–b c_1 sinbx + bc_2 cosbx ] + [ c_1 cosbx + c_2 sin bx] e^(ax) *a

or (dy)/(dx) = e^(ax) [ (bc_2 + ac_1) cos bx + (ac_2 - bc_1 ) sin bx ] .........................(2)

Differentiating both sides of equation (2) with respect to x, we get

(d^2y)/(dx^2) = e^(ax) [ ( b c_2 + ac_1 ) ( - b sin bx) + ( a c_2 - b c_1 ) ( b cos bx) ]

+[bc_2 + a c_1 ) cos bx + ( a c_2 - bc_1 ) sin bx ]

= e^(ax) [ ( a^2c_2 - 2 abc_1 - b^2 c_2 ) sin bx + ( a^2 c_1 + 2 abc_2 - b^2c_1 ) cos bx ]

Substituting the values of (d^2y)/(dx^2) , (dy)/(dx) and y in the given differential equation, we get

L.H.S = a^(ax) [ a^2c_2 - 2 abc_1 - b^2 c_2 ) sin bx + ( a^2c_1 + 2 abc_2 - b^2c_1 ) cos bx ]
\ \ \ \ \ \ - 2 e^(ax)[ ( bc_2 +ac_! ) cos bx + (ac_2 - bc_1 ) sin bx ]
\ \ \ \ \ \ + (a^ +b^2 ) e^(ax) [ c_1 cos bx + c_2 sin bx]

= e^(ax) [ (a^c_2 - 2abc_1 - b^2c_2 - 2 a^2c_2 + 2 abc_1 + a^2 c_2 + b^2 c_2 ) sin bx +
(a^2c_1 + 2 abc_2 - b^2c_1 - 2abc_2 - 2 a^2c_1 + a^2 c_1 + b^2c_1 ) cos bx ]

= e^(ax) [ 0xx sin bx + 0 cos bx ] = e^(ax) xx 0 = 0 RHS

Hence, the given function is a solution of the given differential equation.