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
has a unique solution in some interval of the form -
ii) The orthogonal trajectories of all the parabolas with vertices at the origin and foci on the
is
iii) The normal form of the differential equation
where
iv The solution of the pde is
Answer :
Word Count : 466
Let's address each statement one by one: ### i) The initial value problem: \[ \frac{dy}{dx} = x^2 + y^2, \quad y(0) = 0 \] This statement asks whether there exists a unique solution to this initial value problem in some interval. Answer: False. The differential equation \(\frac{dy}{dx} = x^2 + y^2\) does not satisfy the conditions for the existence and uniqueness of a solution according to the Picard-Lindelöf theorem (which is based on the Lipschitz continuity of the function with respect to \(y\)). Here, \(f(x, y) = x^2 + y^2\) is not Lipschitz continuous in \(y\) because it grows quadratically in \(y\), which violates the condition for uniqueness. Therefore, this problem may have multiple solutions (or ______ ___ ____ _________ _____ _______ _________.
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Let's address each statement one by one: ### i) The initial value problem: \[ \frac{dy}{dx} = x^2 + y^2, \quad y(0) = 0 \] This statement asks whether there exists a unique solution to this initial value problem in some interval. Answer: False. The differential equation \(\frac{dy}{dx} = x^2 + y^2\) does not satisfy the conditions for the existence and uniqueness of a solution according to the Picard-Lindelöf theorem (which is based on the Lipschitz continuity of the function with respect to \(y\)). Here, \(f(x, y) = x^2 + y^2\) is not Lipschitz continuous in \(y\) because it grows quadratically in \(y\), which violates the condition for uniqueness. Therefore, this problem may have multiple solutions (or ______ ___ ____ _________ _____ _______ _________.
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Get Full Answer on WhatsApp
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