`color {red} ♦` Linear differential equations

`color {red} ♦` Steps involved to solve first order linear differential equation

`color {red} ♦` Miscellaneous Examples

`color {red} ♦` Steps involved to solve first order linear differential equation

`color {red} ♦` Miscellaneous Examples

`\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.

`(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.

(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}`

(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

Class 12 Chapter 9 Example 19

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

Class 12 Chapter 9 Example 20

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

Class 12 Chapter 9 Example 21

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

gives that y = 0 when `x = pi/2`

Class 12 Chapter 9 Example 22

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

Class 12 Chapter 9 Example 23

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

`(d^2y)/(dx^2) - 2 a (dy)/(dx) + ( a^2 + b^2) y = 0`

Class 12 Chapter 9 Example 24

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.