PYQ on Ordinary Differential Equation of Higher Order

In examinations, questions from Ordinary Differential Equations (ODEs) of higher order are often asked in the form of solving second or higher-order linear differential equations with constant or variable coefficients.

Typical methods include the use of auxiliary equations, particular integrals, change of variable, reduction of order, and solving simultaneous linear differential equations, unlike first-order equations, which involve derivatives of order two or more, representing systems with acceleration, oscillation, or more complex behavior.

Short Questions on Ordinary Differential Equations of Higher Order

Question 1: Solve - d²y​/dx² − 3dy/dx​ + 2y = 0

Question 2: Find the general solution of d³y/dx³ − y = 0

Question 3: Solve using the method of variation of parameters: y′′+ y = sec⁡x

Question 4: Solve the Cauchy–Euler equation: x²y′′ + 3xy′ + y = 0x²y'' + 3xy' + y = 0x2y′′ + 3xy′ + y = 0.

Question 5: Solve the simultaneous differential equations: dx/dt = 4x + y, dy/dt = −2x + y

Question 6: Solve the non-homogeneous equation: d²y/dx² − y = e^x

Question 7: Find the complementary function (C.F.) and particular integral (P.I.) of: (D² − 2D + 1)y = e^2x

Question 8: Find the particular integral of d²y/dx² + 4y = sin⁡2x.

Question 9: Solve: d²y/dx² + 9y = 0

Question 10: Solve: d³y/dx³− 3d²y/dx²+ 3dy/dx − y = 0

Long Questions on Ordinary Differential Equations of Higher Order

Question 11: Solve (D²+4D + 13) y = 4e^−2xsin⁡3x where D = d/dx.

Question 12: Solve the differential equation y′′′− y′ = 2cos⁡x.

Check if you were right - full answer with solution below.

Short Question on ODEs: Answers

Solution 1:

This is a second-order linear differential equation with constant coefficients.

For a differential equation of the form:

y′′+ ay′+ by = 0

The auxiliary equation is:

m² + am + b = 0

Here, a = −3, b = 2. So:

m² − 3m + 2 = 0
(m − 1)(m − 2) = 0 
 m = 1or m = 2

For distinct real roots m₁ and m₂​, the general solution is:

y(x) = C₁​e^m1​x+C₂​e^m2​x

Substitute m₁ =1, m₂ = 2:

y(x) = C₁e^x + C₂e^2x

Solution 2:

m³ − 1 = 0

Factor using difference of cubes:

m³ − 1 = (m − 1)(m² + m +1) = 0

m = 1, m = −1 ± i√3​​/2

y = C₁​e^x + e^−x/2(C_2​cos^{√3​​x/2}+ C₃​sin^{√3​​x/2})

Solution 3:

y_c​ = C_1​cosx + C₂​sinx

For y′′+ y = f(x), P.I.:

y_p = -y_1 \int \frac{y_2 f(x)}{W} dx + y_2 \int \frac{y_1 f(x)}{W} dx

Here, y₁ = cos⁡x, y₂ = sin⁡x, Wronskian W = 1

y_p = \frac{1}{2} \sin x \ln |\sec x + \tan x|

y = C_1​cosx + C_2​sinx + 1/2​sinxln∣secx + tanx∣​

Solution 4:

Assume solution y = x^m

y′ = mx^m−1,y′′ = m(m − 1)x^m−2

Substitute into the equation:

x²[m(m − 1)x^{m − 2}] + 3x[mx^{m − 1}] + x^m = 0

m² + 2m + 1 = 0

⇒ (m + 1)² = 0

Repeated root m = − 1

y = C_1​x^-1 + C_2​ln x

Solution 5:

\frac{d}{dt}\begin{bmatrix}x \\y\end{bmatrix}=\begin{bmatrix}4 & -2 \\1 & 1\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}

\begin{vmatrix}4 - \lambda & -2 \\1 & 1 - \lambda\end{vmatrix}

= (4 − λ)(1 − λ) + 2 = 0

λ² − 5λ + 6 =0

λ = 2, 3

Eigenvectors:

For λ = 2:

\begin{bmatrix}2 & 1 \\-2 & 2\end{bmatrix}\;\Rightarrow\;v_1 = \begin{bmatrix}1 \\-2\end{bmatrix}

\begin{bmatrix}1 & 1 \\-2 & -2\end{bmatrix}\;\Rightarrow\;v_2 =\begin{bmatrix}1 \\-1\end{bmatrix}

\begin{bmatrix}x \\y\end{bmatrix} = C₁ e^2t \begin{bmatrix}-1 \\2\end{bmatrix} + C₂ e^3t \begin{bmatrix}1 \\-1 \end{bmatrix}

Solution 6 :

Auxiliary equation: m² − 1 = 0⇒ m = ±1

C.F. =C₁e^x+C₂e^−x

Since RHS e^x is already part of CF (repeated root case):

P.I. = Axe^x

Substitute in LHS:

(D² − 1)(Axe^x) = e^x(2A)

To match e^x, 2A = 1
⇒A = 1/ 2

y = C₁ e^x + C₂ e^{- x} + 1/2 xe^x

Solution 7:

Auxiliary equation: (m−1)² = 0

m = 1,1

C.F. = (C₁ + C₂x)e^x

For P.I:

P.I = \frac{e^2x}{(2^2)-2(2) + 1}

\frac{e^2x}{1}

= e^2x

( C₁+ C₂ x)e^x+ e^2x

Solution 8:

The homogeneous equation y′′+ 4y = 0 has solutions cos⁡2x, sin⁡2x. RHS sin⁡2x is a solution of the homogeneous equation → resonance. Try a particular solution of the form

y_p ​= x(Acos2x+Bsin2x).

y′_p​ = Acos2x +Bsin2x + x(−2Asin2x + 2Bcos2x).

y′′_p = −2Asin⁡2x + 2Bcos⁡2x + (−2Asin⁡2x+2Bcos⁡2x) + x(−4Acos⁡2x−4Bsin⁡2x)
= −4Asin⁡2x + 4Bcos⁡2x + x(−4Acos⁡2x − 4Bsin⁡2x)
= -4Asin2x + 4Bcos2x + x(-4Acos2x - 4Bsin2x)

Now

y′′_p + 4y_p= (−4Asin⁡2x + 4Bcos⁡2x) + x(−4Acos⁡2x − 4Bsin⁡2x + 4Acos⁡2x + 4Bsin⁡2x)

the x-term cancels, so

y′'_p+ 4y_p= 4Bcos⁡2x − 4Asin⁡2x

Set equal to sin⁡2x:

4Bcos⁡2x − 4Asin⁡2x = sin⁡2x.

Thus a particular integral is

y_p= −1/4 xcos⁡2x.

Solution 9:

Characteristic equation r²+9 = 0
⇒r = ±3i. So the general solution is

y = C₁cos⁡3x + C₂sin⁡3x

with arbitrary constants C_1,C₂ .

Solution 10:

d³y/dx³​−3d²y/dx² ​+ 3dy/dx​ − y = 0.

Recognize the left-hand operator as (D−1)³ where D=d/dx. The characteristic equation is (r−1)³= 0 → a triple root r = 1.

So the general solution is

y = (C₁+C₂x+C₃x²)e^x

with constants C_1,C₂,C₃.

Short Question on ODEs: Answers

Solution 11:

Auxiliary equation for the homogeneous operator:

r²+4r+13 = 0

Compute the roots:

r = \frac{-4 \pm \sqrt{16 - 52}}{2} =−2 ± 3i.

y_c​= e⁻²x(C₁​cos3x + C₂​sin3x).

RHS is 4e^−2xsin⁡3x, which matches the homogeneous frequency (resonance). Use the shift trick: write y=e⁻²xv(x). Then

(D²+4D+13)(e − 2xv) =e ^-2x(D² + 9)v

(verify by substituting D → D − 2). So we need (D²+9)v = 4sin⁡3x with v chosen of the form v = x(Acos⁡3x + Bsin⁡3x). For v = x(Acos⁡3x + Bsin⁡3x).

v′′+ 9v = 6Bcos3x − 6Asin3x.

Set equal to 4sin⁡3x4\sin3x4sin3x. Thus

6B = 0
⇒B = 0,
−6A = 4
⇒A = −23.

Hence

v_p = −23xcos⁡3x, y_p = e^−2x v_p = −2/3xe^−2xcos⁡3x

y = e^−2x(C1cos⁡3x +C2sin⁡3x) − 2/3xe^−2xcos⁡3x.

Solution 12:

y′′′−y′=2cosx.

Homogeneous: characteristic r³−r = r(r²−1) = r(r − 1)(r+1)
⇒ roots 0, 1 ,−1

y_c= C₁+C₂e^x+C₃e^−x.

For RHS try y_p = Acos⁡x +Bsin⁡x. Compute

y′_p = −Asin⁡x + Bcos⁡x, y′′′_p = Asin⁡x − Bcos⁡x.

So

y′′′_p− y′_p = (Asin⁡x − Bcos⁡x) − (−Asin⁡x + Bcos⁡x) = 2Asin⁡x − 2Bcos⁡x.

Set equal to 2cos⁡x:

2A = 0
⇒A = 0,  −2B = 2 ⇒ B = −1. So y_p = −sin⁡x.

Hence, y= C₁+ C₂e^x +C₃e^−x − sin⁡x.