Partial Differential Equation MCQ - 2 - Mathematics MCQ

Using Euler theorem 
xf_x + yf_y = n f(x, y) 
Substituting x = 0; n = -96 and y = 1 we have 
fy = -96. f(0, 1) = -96.(1⁄1) 
= – 96.

Tan(y) = Tan(x)

ANSWER :- a

Solution :- Lagrange auxiliary equations are :-

dx/(y^2 + z^2 - x^2) = -dy/(-2xy) = -dz/(2xz)

Taking the last two members, we get

dy/y = dz/z

Integrating log y = logz + logc1

=y/z = c1

Using ,multipliers x,y,z we get (xdx + ydy + zdz)/-x(x^2 + y^2 + z^2) 

(xdx + ydy + zdz)/-x(x^2 + y^2 + z^2) = dx/(-2xz)

2(xdx + ydy + zdz)/(x^2 + y^2 + z^2) = dz/z

Integrating, log(x^2 + y^2 + z^2) = logz + logc2

(x^2 + y^2 + z^2) = zc2

Hence the required solution is f(c1,c2) = 0

= f(y/z, (x^2 + y^2 + z^2)/z) = 0

Given: z = 3xy + 4x² 
Using partial differentiation, we need to differentiate the function z with respect to x keeping y as constant. Thus, ∂z / ∂x = 3y + 8x.

ANSWER :- c

Solution :-  z(xzx−yzy)=y^2−x^2

xzx−yzy = (y^2−x^2)z

dx/dt=x  , letting x(0)=1 , we have x=e^t

dy/dt=−y , letting y(0)=y0 , we have y=y0e^(−t) = y0/x

dz/dt=(y^2−x^2)/z

=y^20e^(−2t) −e^(2t)/z , 

we have z^2=f(y0)−y^20e^(−2t)−e^(2t) 

=f(xy)−y^2−x^2 ,

 i.e. x2+y2+z2=f(xy)

z^2=- x^2 - y^2 + f(xy)

Given: (y’)² + 1 = 0
Consider if y = 2x, then y’ = 2 and hence the left-hand side of the equation becomes 3 which is greater than 1. Therefore, the left-hand side of the equation will always be greater than, or equal to one and thus cannot be zero and hence the differential equation is not satisfied.

We have

Given: x³ + y³ = 3axy
To find the tangent to the curve at the origin, we need to equate the lowest degree term to 0.
Therefore, 3axy = 0, which gives x = 0 and y = 0 as two tangents to the curve at origin.

By the definition of partial derivatives


If we moving along the curve y = mx²

⇒ f(x, y)  is not continuous.

Explanation: The first step is to expand the given equation by replacing every Δy_n by (y_n+1 – y_n). Order of a difference equation is given by, (n+3-n)/1 which is actually 3.

Differentiation of x³ is 3x² 
differentiation of y³ is 3y² dy / dx
differentiation of -3xy is [-3y-3x(dy/dx)]
differentiation of y² is 2y dy / dx
Hence, 

The division rule of differentiation for two functions is given by,

Think of the PDE as ∇u⋅(a(x,y),b(x,y))=0∇u⋅(a(x,y),b(x,y))=0 so that, geometrically it must be that dy/dx = [b(x,y)/a(x,y)] The solution of the ODE defines the characteristic curves of the PDE, along which solutions (to the PDE) are constant. Then u(x,y) = f(C) where the function ff is determined from the given auxiliary condition.

For example, here the ODE is dy/dx = y/x

 so C = ln|y/x| 

Then u(x,y) = f(C) = f(ln|y/x|).

To determine f, apply the given condition: u(1,y) = f(ln|y|) = y

⟹ f(y) = e^y 

Thus, u(x,y) = exp(ln|y/x|)

= |y/x|

To solve this problem let us assume the given function is dependent on α Such that α = 2 & thus 

f (α) = log(α + 1) + c
or f (α) = log(α + 1) …… neglecting constant since the function is assumed
thus f (2) = log(2 + 1) = log(3).