(D² + 1) y = 0

A.E. is (D² + 1) = 0

⇒ D = ± i

Solution is, y = (C₁ cos t + C₂ sin t)

Given that, y(0) = 1

⇒ 1 = (C₁ + 0) ⇒ C₁ = 1

y’ = -C₁ sin t + C₂ cos t

y’(0) = 0

⇒ 0 = 0 + C₂ ⇒ C₂ = 0

⇒ y = cos t

$\Rightarrow y\left(\frac{\pi }{2}\right)=\cos \left(\frac{\pi }{2}\right)=0$

One dimensional wave equation: $\frac{{\partial }^{2}u}{\partial t^2}=C^2\frac{{\partial }^{2}u}{\partial x^2}$

Two-dimensional wave equation: $\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}$

Important:

Heat equation: $\frac{\partial u}{\partial t}=C^2\frac{{\partial }^{2}u}{\partial x^2}$

$y\frac{dy}{dx}=6x^2+5$

⇒ y dy = (6x² + 5) dx

By integrating both the sides, we get

$\Rightarrow \frac{y^2}{2}=2x^3+5x+C$ 

⇒ y² = 4x³ + 10x + C

y(0) = 2

⇒ C = 4

Now, y² = 4x³ + 10x + 4

At x = 1, y² = 4 + 10 + 4 = 18

$\Rightarrow y\left(1\right)=\sqrt{18}=\pm 3\sqrt{2}$

Explanation:

$\frac{{\partial }^2}{\partial y^2}+z=0$

If z were function of y alone, the solution would have been z = A sin y + B cos y, where A and B are constants.

Since z is a function of x and y, A and B can be arbitrary functions of x.

Hence the solution of the given equations is,

z = f(x) sin y + ϕ(x) cos y

$\frac{\partial z}{\partial y}=f\left(x\right)\cos y-\varphi \left(x\right)\sin y$

At, y = 0, z = e^x

⇒ e^x = ϕ(x)

At, $y=0,\frac{\partial z}{\partial y}=e^{-x}$

⇒ e^-x = f(x)

The solution is,

Linear differential equations with variable coefficients can be reduced to linear differential equations with constant coefficients by suitable substitutions.

Euler Cauchy Homogeneous linear equation:

$x^n\frac{d^{n}y}{d{x}^n}+c_1{x}^{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}+...+c_{n-1}x\frac{dy}{dx}+c_{n}y=f(x)$

Take x = et ⇒ t = log x

$\begin{array}{l}\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{dy}{dt}.\frac{1}{x}\Rightarrow x\frac{dy}{dx}=\frac{dy}{dt}=Dy\\ x^2\frac{d^{2}y}{d{x}^2}=D(D-1)y;x^3\frac{d^{3}y}{d{x}^3}=D(D-1)(D-2)y\end{array}$

Given that, x, x ln x. and ln²x as linearly independent solutions.

y = C₁x + C₂x ln x + C₃x ln² x

put, t = ln x

⇒ x = e^t

⇒ y = C₁ e^t + C₂ t e^t + C₃ t² e^t

⇒ y = (C₁ + C₂t + C₃t²) e^t

From the above equation, roots of auxiliary equation are, D = 1, 1, 1

Now, the auxiliary equation is,

(D - 1) (D - 1) (D - 1) = 0

⇒ (D² + 1 - 2D) (D - 1) = 0

⇒ D³ - 3D² + 3D - 1 = 0      ----(1)

The standard Cauchy’s differential equation is in the form of

K₁ D(D - 1) (D - 2) + K₂ D(D - 1) + K₃ D + K₄ = 0

⇒ K₁ [D³ - 3D² + 2D] + K₂ [D² - D] + K₃D + K₄ = 0      ----(2)

By comparing the above two equations,

K₁ = 1, K₄ = -1

-2K₁ + K₂ = -3 ⇒ K₂ = 0

⇒ K₃ = 1

Now the equation (2) becomes,

D(D - 1) (D - 2) + D - 1 = 0

⇒ D(D - 1) (D - 2) y + Dy - y = 0

$\Rightarrow x^3\frac{d^{3}y}{d{x}^3}+x\frac{dy}{dx}-y=0$

⇒ x³y''' + xy' - y = 0

$\frac{dy}{dx}=\frac{y^2}{x}$

$\Rightarrow \frac{dy}{y^2}=\frac{dx}{x}$

$=\int \frac{dy}{y^2}=\int \frac{dx}{x}$

$\Rightarrow -\frac{1}{y}=\ln \left(x\right)+c$

put y(1) = 1

$\Rightarrow -\frac{1}{\left(1\right)}=\ln \left(1\right)+c$

⇒ c = - 1

$\therefore -\frac{1}{y}=\ln \left(x\right)-1$

$\Rightarrow y=\frac{1}{1-\ln \left(x\right)}$

For y to be finite

$1-ln\left(x\right)\ne 0andx>0$

⇒ ln (x) ≠ 1

⇒ x ≠ e¹

⇒ x ≠ e

⇒ x ≠ 2.718

So the correct range of x is 0 < x < 2.71 and 2.718 < x ≤ ∞

From the given options, 3 ≤ x ≤ ∞ satisfies the above conditions.

Concept:

Euler Cauchy Homogeneous linear equation:

$x^n\frac{d^{n}y}{d{x}^n}+c_1{x}^{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}+...+c_{n-1}x\frac{dy}{dx}+c_{n}y=f(x)$

Take x = et ⇒ t = log x

$\begin{array}{l}\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{dy}{dt}.\frac{1}{x}\Rightarrow x\frac{dy}{dx}=\frac{dy}{dt}=Dy\\ x^2\frac{d^{2}y}{d{x}^2}=D(D-1)y;x^3\frac{d^{3}y}{d{x}^3}=D(D-1)(D-2)y\end{array}$

Calculation:

$\left(x^2{D}^2-2\right)y=x^2+\frac{1}{x}$

$\Rightarrow \left(D\left(D-1\right)-2\right)y=e^{2t}+\frac{1}{e^t}$

⇒ (D² – D - 2)y = e^2t + e^-t

$PI=\frac{1}{\left(D^{2}D-2\right)}\left(e^{2t}+e^{-t}\right)$ 

$=\frac{1}{\left(D^2-D-2\right)}{e}^{2t}+\frac{1}{\left(D^2-D-2\right)}{e}^{-t}$ 

$=\frac{t}{2D-1}{e}^{2t}+\frac{t}{2D-1}{e}^{-t}$ 

$=\frac{t}{3}{e}^{2t}+\frac{t}{-3}{e}^{-t}$  

$=\frac{t}{3}\left(e^{2t}-e^{-t}\right)=\frac{lnx}{3}\left(x^2-\frac{1}{x}\right)$ 

Concept:

Euler Cauchy Homogeneous linear equation:

$x^n\frac{d^{n}y}{d{x}^n}+c_1{x}^{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}+...+c_{n-1}x\frac{dy}{dx}+c_{n}y=f(x)$

Take x = et ⇒ t = log x

$\begin{array}{l}\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{dy}{dt}.\frac{1}{x}\Rightarrow x\frac{dy}{dx}=\frac{dy}{dt}=Dy\\ x^2\frac{d^{2}y}{d{x}^2}=D(D-1)y;x^3\frac{d^{3}y}{d{x}^3}=D(D-1)(D-2)y\end{array}$

Calculation:

$x^2\frac{d^{3}y}{d{x}^3}-4x\frac{d^{2}y}{d{x}^2}+6\frac{dy}{dx}=4$

By multiplying with ‘x’ on both sides

$\Rightarrow x^3\frac{d^{3}y}{d{x}^3}-4x^2\frac{d^{2}y}{d{x}^2}+6x\frac{dy}{dx}=4x$

The given equations become,

⇒ [D (D – 1) (D – 2) – 4 D (D – 1) + 6D] y = 4x

⇒ (D³ – 3D² + 2D – 4D² + 4D + 6D) y = 4x

⇒ (D³ – 7D² + 12D) y = 4x

Auxiliary equation: D³ – 7D² + 12D = 0

⇒ D (D² – 7D + 12) = 0

⇒ D = 0, 3, 4

Complementary solution (CF) = C₁ + C₂ e^3t + C₃ e^4t

CF = C1 + C2 x³ + C₃x⁴  

Concept:

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

Calculation:

$x^2\frac{d^{2}y}{d{x}^2}-7x\frac{dy}{dx}+16y=0$

Put x = e^t

⇒ t = ln x

$\begin{array}{l}\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{dy}{dt}.\frac{1}{x}\Rightarrow x\frac{dy}{dx}=\frac{dy}{dt}=Dy\\ x^2\frac{d^{2}y}{d{x}^2}=D(D-1)y;x^3\frac{d^{3}y}{d{x}^3}=D(D-1)(D-2)y\end{array}$

Now, the above differential equation becomes

D (D – 1) y – 7 D y + 16 y = 0

⇒ D² y – D y – 7 D y + 16 y = 0

⇒ (D² – 8 D + 16) y = 0

Auxiliary equation:

(D² – 8 D + 16) = 0

⇒ D = 4

The solutions for the above roots of auxiliary equations are:

y(t) = (c₁ + c₂ t) e^4t

The given partial differential equation can be written as

(4D² + 12DD’ + 9D’²) z = 0

Its auxiliary equation is,

4m² + 12m + 9 = 0

$\Rightarrow m=\frac{-3}{2},\frac{-3}{2}$ 

The complete solution is,

z = f₁ (y – 1.5 x) + x f₂ (y – 1.5 x)