Unit 1 | unit 1 higher order linear differential equations and applications

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Unit - 1

Higher order linear differential equations and applications

Q1) Solve

A1) Its auxiliary equation is-

Where-

Therefore the complete solution is-

Q2) What is inverse operator?

A2)

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

Q3) Find the P.I. Of (D + 2)

A3)

P.I. =

Now we will evaluate each term separately-

And

Therefore-

Q4) Solve (D – D’ – 2 ) (D – D’ – 3) z =

A4)

The C.F. Will be given by-

Particular integral-

Therefore the complete solution is-

Q5) Find the P.I. Of

A5)

Q6) Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

A6)

Q7) Find P.I. Of

A7)

P.I =

Replace D by D+1

Put

Q8) Solve

A8)

Here first we will find the C.F.-

Its auxiliary equation will be-

Here we get-

Now we will find P.I.-

Now the complete solution is-

Q9) Solve the following DE by using variation of parameters-

A9)

We can write the given equation in symbolic form as-

To find CF-

It’s A.E. Is

So that CF is-

To find PI-

Here

Now

Thus PI =

So that the complete solution is-

Q10) Solve the following by using the method of variation of parameters.

A10)

This can be written as-

C.F.-

Auxiliary equation is-

So that the C.F. Will be-

P.I.-

Here

Now

Thus PI =

So that the complete solution is-

Q11) Solve

A11)

The Auxiliary equation is

The C.F is

P.I

The Complete Solution is

Q12) Solve

A12)

The auxiliary equation is

The C.F is

But

The Complete Solution is

Q13) Solve (4D² +4D -3)y =

A13)

Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3)(2m – 1) = 0

m = ,

Complementary function: CF is A+ B

Now we will find particular integral,

P.I. = f(x)

= .

= . = .

General solution is y = CF + PI

= A+ B .

Q14) Solve

A14)

Given,

Here Auxiliary equation is

Q15) Solve

A15)

Auxiliary equation are

Q16)

A16)

Auxiliary equation are

Q17) Solve

A17)

The AE is

Complete solution = CF + PI

Q18) Find the PI of

A18)

Q19) Solve

A19)

The AE is

We know,

Complete solution is y= CF + PI

Q20)

A20)

Putting,

AE is

CS = CF + PI

Q21) Finding the optimal current of an electrical circuit(RL circuits) in which the initial

Condition is i=0 at t=0

$C:\Users\Acer\Documents\electrical circuit MODIFIED.jpg$

A21)

By Kirchhoff voltage law(KVL) method, we get

The differential equation for the RL circuit will be

In which initial conditions are i=0 at t=0

The standard form of the equation is,

Dividing the differential equation by L to obtain

The integrating factor is

Multiplying the above equation with standard form gives rise to

By applying integration on both sides we get

Now applying i=0 at t-0 gives us

C=-

NOW

= t

Therefore by finding current of the RL circuits is i=

Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.

Unit - 1

Higher order linear differential equations and applications

Q1) Solve

A1) Its auxiliary equation is-

Where-

Therefore the complete solution is-

Q2) What is inverse operator?

A2)

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

Q3) Find the P.I. Of (D + 2)

A3)

P.I. =

Now we will evaluate each term separately-

And

Therefore-

Q4) Solve (D – D’ – 2 ) (D – D’ – 3) z =

A4)

The C.F. Will be given by-

Particular integral-

Therefore the complete solution is-

Q5) Find the P.I. Of

A5)

Q6) Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

A6)

Q7) Find P.I. Of

A7)

P.I =

Replace D by D+1

Put

Q8) Solve

A8)

Here first we will find the C.F.-

Its auxiliary equation will be-

Here we get-

Now we will find P.I.-

Now the complete solution is-

Q9) Solve the following DE by using variation of parameters-

A9)

We can write the given equation in symbolic form as-

To find CF-

It’s A.E. Is

So that CF is-

To find PI-

Here

Now

Thus PI =

So that the complete solution is-

Q10) Solve the following by using the method of variation of parameters.

A10)

This can be written as-

C.F.-

Auxiliary equation is-

So that the C.F. Will be-

P.I.-

Here

Now

Thus PI =

So that the complete solution is-

Q11) Solve

A11)

The Auxiliary equation is

The C.F is

P.I

The Complete Solution is

Q12) Solve

A12)

The auxiliary equation is

The C.F is

But

The Complete Solution is

Q13) Solve (4D² +4D -3)y =

A13)

Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3)(2m – 1) = 0

m = ,

Complementary function: CF is A+ B

Now we will find particular integral,

P.I. = f(x)

= .

= . = .

General solution is y = CF + PI

= A+ B .

Q14) Solve

A14)

Given,

Here Auxiliary equation is

Q15) Solve

A15)

Auxiliary equation are

Q16)

A16)

Auxiliary equation are

Q17) Solve

A17)

The AE is

Complete solution = CF + PI

Q18) Find the PI of

A18)

Q19) Solve

A19)

The AE is

We know,

Complete solution is y= CF + PI

Q20)

A20)

Putting,

AE is

CS = CF + PI

Q21) Finding the optimal current of an electrical circuit(RL circuits) in which the initial

Condition is i=0 at t=0

$C:\Users\Acer\Documents\electrical circuit MODIFIED.jpg$

A21)

By Kirchhoff voltage law(KVL) method, we get

The differential equation for the RL circuit will be

In which initial conditions are i=0 at t=0

The standard form of the equation is,

Dividing the differential equation by L to obtain

The integrating factor is

Multiplying the above equation with standard form gives rise to

By applying integration on both sides we get

Now applying i=0 at t-0 gives us

C=-

NOW

= t

Therefore by finding current of the RL circuits is i=

Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.

Unit - 1

Higher order linear differential equations and applications

Q1) Solve

A1) Its auxiliary equation is-

Where-

Therefore the complete solution is-

Q2) What is inverse operator?

A2)

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

Q3) Find the P.I. Of (D + 2)

A3)

P.I. =

Now we will evaluate each term separately-

And

Therefore-

Q4) Solve (D – D’ – 2 ) (D – D’ – 3) z =

A4)

The C.F. Will be given by-

Particular integral-

Therefore the complete solution is-

Q5) Find the P.I. Of

A5)

Q6) Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

A6)

Q7) Find P.I. Of

A7)

P.I =

Replace D by D+1

Put

Q8) Solve

A8)

Here first we will find the C.F.-

Its auxiliary equation will be-

Here we get-

Now we will find P.I.-

Now the complete solution is-

Q9) Solve the following DE by using variation of parameters-

A9)

We can write the given equation in symbolic form as-

To find CF-

It’s A.E. Is

So that CF is-

To find PI-

Here

Now

Thus PI =

So that the complete solution is-

Q10) Solve the following by using the method of variation of parameters.

A10)

This can be written as-

C.F.-

Auxiliary equation is-

So that the C.F. Will be-

P.I.-

Here

Now

Thus PI =

So that the complete solution is-

Q11) Solve

A11)

The Auxiliary equation is

The C.F is

P.I

The Complete Solution is

Q12) Solve

A12)

The auxiliary equation is

The C.F is

But

The Complete Solution is

Q13) Solve (4D² +4D -3)y =

A13)

Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3)(2m – 1) = 0

m = ,

Complementary function: CF is A+ B

Now we will find particular integral,

P.I. = f(x)

= .

= . = .

General solution is y = CF + PI

= A+ B .

Q14) Solve

A14)

Given,

Here Auxiliary equation is

Q15) Solve

A15)

Auxiliary equation are

Q16)

A16)

Auxiliary equation are

Q17) Solve

A17)

The AE is

Complete solution = CF + PI

Q18) Find the PI of

A18)

Q19) Solve

A19)

The AE is

We know,

Complete solution is y= CF + PI

Q20)

A20)

Putting,

AE is

CS = CF + PI

Q21) Finding the optimal current of an electrical circuit(RL circuits) in which the initial

Condition is i=0 at t=0

$C:\Users\Acer\Documents\electrical circuit MODIFIED.jpg$

A21)

By Kirchhoff voltage law(KVL) method, we get

The differential equation for the RL circuit will be

In which initial conditions are i=0 at t=0

The standard form of the equation is,

Dividing the differential equation by L to obtain

The integrating factor is

Multiplying the above equation with standard form gives rise to

By applying integration on both sides we get

Now applying i=0 at t-0 gives us

C=-

NOW

= t

Therefore by finding current of the RL circuits is i=

Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.

Unit - 1

Higher order linear differential equations and applications

Q1) Solve

A1) Its auxiliary equation is-

Where-

Therefore the complete solution is-

Q2) What is inverse operator?

A2)

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

Q3) Find the P.I. Of (D + 2)

A3)

P.I. =

Now we will evaluate each term separately-

And

Therefore-

Q4) Solve (D – D’ – 2 ) (D – D’ – 3) z =

A4)

The C.F. Will be given by-

Particular integral-

Therefore the complete solution is-

Q5) Find the P.I. Of

A5)

Q6) Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

A6)

Q7) Find P.I. Of

A7)

P.I =

Replace D by D+1

Put

Q8) Solve

A8)

Here first we will find the C.F.-

Its auxiliary equation will be-

Here we get-

Now we will find P.I.-

Now the complete solution is-

Q9) Solve the following DE by using variation of parameters-

A9)

We can write the given equation in symbolic form as-

To find CF-

It’s A.E. Is

So that CF is-

To find PI-

Here

Now

Thus PI =

So that the complete solution is-

Q10) Solve the following by using the method of variation of parameters.

A10)

This can be written as-

C.F.-

Auxiliary equation is-

So that the C.F. Will be-

P.I.-

Here

Now

Thus PI =

So that the complete solution is-

Q11) Solve

A11)

The Auxiliary equation is

The C.F is

P.I

The Complete Solution is

Q12) Solve

A12)

The auxiliary equation is

The C.F is

But

The Complete Solution is

Q13) Solve (4D² +4D -3)y =

A13)

Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3)(2m – 1) = 0

m = ,

Complementary function: CF is A+ B

Now we will find particular integral,

P.I. = f(x)

= .

= . = .

General solution is y = CF + PI

= A+ B .

Q14) Solve

A14)

Given,

Here Auxiliary equation is

Q15) Solve

A15)

Auxiliary equation are

Q16)

A16)

Auxiliary equation are

Q17) Solve

A17)

The AE is

Complete solution = CF + PI

Q18) Find the PI of

A18)

Q19) Solve

A19)

The AE is

We know,

Complete solution is y= CF + PI

Q20)

A20)

Putting,

AE is

CS = CF + PI

Q21) Finding the optimal current of an electrical circuit(RL circuits) in which the initial

Condition is i=0 at t=0

$C:\Users\Acer\Documents\electrical circuit MODIFIED.jpg$

A21)

By Kirchhoff voltage law(KVL) method, we get

The differential equation for the RL circuit will be

In which initial conditions are i=0 at t=0

The standard form of the equation is,

Dividing the differential equation by L to obtain

The integrating factor is

Multiplying the above equation with standard form gives rise to

By applying integration on both sides we get

Now applying i=0 at t-0 gives us

C=-

NOW

= t

Therefore by finding current of the RL circuits is i=

Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.

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