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  • Form the partial differential equation by eliminating the achitrary constants A and B from 2-4e sin Bx

Form the partial differential equation by eliminating the achitrary constants A and B from 2-4e sin Bx

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Given- $F(x, A, B) = 2 - 4e \sin(Bx)$

Solution- First differentiate $F(x, A, B) = 2 - 4e \sin(Bx)$ with respect to $B$:  $\frac{\partial F}{\partial B} = -4e x \cos(Bx)$.
The differentiate with respect to $x$:  $\frac{\partial F}{\partial x} = 4e B \cos(Bx)$.
Differentiate again with respect to $x$:  $\frac{\partial^2 F}{\partial x^2} = -4e B^2 \sin(Bx)$.
Now, substitute the results into the equation: $\frac{\partial^2 F}{\partial x^2} + F = 0$.
Thus, the partial differential equation formed is:  $\frac{\partial^2 F}{\partial x^2} + F = 0$.

Posted by

Saniya Khatri

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