Consider the following partial differential equaiton 3∂2ϕ∂x2+B∂2ϕ∂x∂y+3∂2ϕ∂y2+4ϕ=0
For the equation to be classified as parabolic, the value of B2 must be
36
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Solution
The correct option is A 36 3∂2ϕ∂x2+B∂2ϕ∂x∂y+3∂ϕ∂y2+4ϕ=0
Compare A∂2ϕ∂x2+B∂2ϕ∂x∂y+C∂ϕ∂y2+Dϕ=0 A=3,B=?;C=3
Equation to be parabolic B2−4AC=0 B2−4×3×3=0 B2=36