IGNOU MTE 8 2026 - . State whether the following statement are true or false. Justify you

Question

. State whether the following statement are true or false. Justify your answer with the help of a short proof or a counter-example.
i) The initial value problem

$$\frac{dy}{dx} = x^2 + y^2, \quad y(0) = 0$

has a unique solution in some interval of the form -h < x < h.
ii) The orthogonal trajectories of all the parabolas with vertices at the origin and foci on the x-axis is $x^2 + 2y^2 = c^2$ .
iii) The normal form of the differential equation

$$\frac{d^2v}{dx^2} + v = -3 \sin 2x, \text{ where } v = ye^{-x^2}.$
iv) The solution of the pde $\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = z^2$ is $z = -[y + f(x - y)]^{-1}$ . (Note: The image appears to have a small typo in the solution provided, it should likely be to the power of -1 based on standard solutions, however, transcribing exactly as seen: $z = -[y + f(x - y)]$ .)
v) The pde $u_{xx} + x^2 u_{xy} - \left( \frac{x^2}{2} + \frac{1}{4} \right) u_{yy} = 0$ is hyperbolic in the entire xy-plane.

09 Jan 2026

Answer :

Word Count : 381

i) Consider the initial value problem: [ \frac{dy}{dx} = x^2 + y^2, \quad y(0) = 0. ] The function (f(x,y) = x^2 + y^2) and its partial derivative with respect to (y), (f_y = 2y), are both continuous everywhere. By the Picard-Lindelöf theorem, this guarantees the existence and uniqueness of a solution in some interval around (x=0). True. --- ii) The parabolas with vertices at the origin and foci on the x-axis are of the form: [ x^2 + 2y^2 = c^2. ] To find orthogonal trajectories, replace (\frac{dy}{dx}) by (-\frac{dx}{dy}). The slope of the given family is: [ \frac{dy}{dx} = -\frac{x}{2y} \implies \text{slope of orthogonal trajectories} = \frac{2y}{x}. ]

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