Higher Order Linear Differential Equations with Constant Coefficients MCQ Quiz in తెలుగు - Objective Question with Answer for Higher Order Linear Differential Equations with Constant Coefficients - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

For the given differential equation:

(D² + (p + q)D + pq)z = 0

⇒ f(D)z = 0

Where f(D) = D² + (p + q)D + pq

Consider in terms of M ⇒ f(m) = 0

M² + (p + q)m + pq = 0

M₁ = -q & m₂ = -p

Hence, its solution is

Z = C₁e^-pt + C₂e^-qt

Concept:

Given equation is

This is a homogeneous second order differential equation,

So (D² + 16)y = 0

D² = m²

⇒ m² + 16 = 0     ⇒ m = ± 4i = 0 ± 4i

Solution is given as in this case roots are complex, m = α ± i β

y = (C₁ cos βx + C₂ sin βx) e^αx

= (C₁ cos 4x + C₂ sin 4x) e^ox = C₁ cos 4x + C₂ sin 4x

Now y’ = -4C₁ sin 4x + 4C₂ cos 4x

Applying Boundary condition,

y’ (0) = 1  ⇒ -4C₁ sin (0) + 4C₂ cos(0) = 1

Putting another boundary condition.

So this equation has no solution.

Explanation:

y” – 4y’ + 3y = 2t – 3t²

f(D) = D² – 4D + 3

= – 2 – 2t – t²

Explanation:

Above given equation is a linear differential equation of order = 2. Such equations are solved using CF + PI method.

Let 

⇒ D²x – 5Dx + 6x = 0

⇒ (D² – 5D + 6)x = 0

Thus auxiliary equation (obtained by replacing D with m) is m² – 5D + 6 = 0.

Roots of above obtained auxiliary equation are-

m² – 5D + 6 = 0

⇒ (m – 2) (m - 3) = 0

⇒ m₁ = 2 or m₂ = 3

When both roots of auxiliary equation are real and distinct

Thus general solution of above D.E is

Applying boundary conditions, x(0) = 0

⇒ O = C₁ + C₂

⇒ C₁ = -C₂     ---(i)

Using initial condition 

⇒ 10 = 2C₁ + 3C₂       ---(ii)

Solving equation (i) & (ii) simultaneously,

C₁ = -10 and C₂ = 10

Thus, x = -10e^2t + 10e^3t

Concept:

The most convenient way to solve a first-order linear differential equation is by implementing the “integrating factor approach”. To apply the “integrating factor approach”, the equation must be expressed as , where P and Q are either function of x or constants. Then the integrating factor (I.F.) can be given by  and the solution of the differential equation is given by,  where c is an integrating constant.

If a linear first-order differential equation is not given in form of , then if possible, it’s better to bring it to that form either by substituting or rearranging.

Calculation:

Given the differential equation,  which is not in the  form.

Put, 

Hence the equation reduces to,  which is now in the  form.

Hence, the solution is 

  Hence, by replacing t we get, 

Given, 

The final form of the solution is, 

Mistake Point:

Make sure to change the argument of trigonometric function into the “radian” of “degree” in the calculator. In general, by default, the argument of a trigonometric function is set to “degree” in online Calculator. 

     ----(1)

There are two constants (a, b) so order of differential equations will be of order two.

       ----(2)

       ----(3)

 which is free from the arbitrary constants a and b and hence this the required differential equation.

Given that,

By differentiating with respect to ‘t’

AE equitation is,

(D² + 1) = 0

⇒ D = ± i

The solution for x is,

x(t) = c₁ cos t + c₂ sin t       ----(1)

        ----(2)

Differentiate with respect to ‘t’

AE equation is,

(D² + 1) is,

⇒ D = ± i

The solution for y is,

y(t) = c₃ cos t + c₄ sin t        ----(3)

From equation (2)

⇒ -x(t) = -c₃ sin t + c₄ cos t

⇒ -c₁ cos t – c₂ sin t = -c₃ sin t + c₄ cos t

By comparing both sides,

c₂ = c₃, c₄ = -c₁

The equations (1) and (3) becomes,

x(t) = c₁ cos t + c₂ sin t        ----(4)

y(t) = c₂ cos t – c₁ sin t        ----(5)

The given initial conditions are,

x(0) = 1, y(0) = 1

⇒ 1 = c₁ cos 0 + c₂ sin 0 ⇒ c₁ = 1

And

1 = c₂ cos 0 – c₁ sin 0 ⇒ c₂ = 1

Now, the equations (4) and (5) becomes,

x(t) = cos t + sin t

y(t) = cos t – sin t

x(π/2) = 1, y(π/2) = -1

x(π/2) + y(π/2) = 1 – 1 = 0

By applying the Laplace transform,

By applying the inverse Laplace transform

y(1) = 3 e^-1

⇒ y(1) = e^-1 + (1 + y’(0)) e^-1 = 3 e^-1

⇒ y’(0) = 1

Now the equation of y(t) becomes

y(t) = (1 + 2t) e^-t

At t = 2,

y(2) = 5 e^-2

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

The given differential equation is, 

D² + 5D + 6 = 0

Auxiliary equation: m² + 5m + 6

⇒ m = -2, -3

The solution is, y = C₁ e^-2x + C₂ e^-3x

y(0) = 2

⇒ C₁ + C₂ = 2

⇒ -2C₁ – 3C₂ = 0

By solving the above two equations, C₁ = 6 and C₂ = -4

Now, the solution is, y(x) = 6 e^-2x – 4 e^-3x

At x = 1, 

Concept:

The auxiliary equation is of the form

D² + 4 = 0

D² = -4

D = ± i2

There are complex roots

The general solution of differential equations is:

y = C₁ cos 2x + C₂ sin 2x

The boundary conditions

y(0) = C₁.1 + C₂.(0) = 0

⇒ C1 = 0

Also, , i.e.

Solving this we get, C₂ = 2

The solution of the differential Equation is, therefore:

y = 2 sin 2x