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Engineering Mathematics Question on Differential Equations

Consider a system of the following two partial differential equations :
∂α∂x=−2∂β∂t\frac{∂\alpha}{∂x}=-2\frac{∂\beta}{∂t}
∂β∂x=−2∂α∂t\frac{∂\beta}{∂x}=-2\frac{∂\alpha}{∂t}
Which one of the following choices is a possible solution for the system ?

A

α(t, x) = (x - t)2 + (x + t)2 and β(t, x) = (x - t)2 - (x + t)2.

B

α(t, x) = (x - 2t)2 + (x +2t)2 and β(t, x) = (x - 2t)2 - (x + 2t)2.

C

α(t,x)=(x−t2)2+(x+t2)2 and β(t,x)=(x−t2)2−(x+t2)2\alpha(t,x)=(x-\frac{t}{2})^2+(x+\frac{t}{2})^2\ \text{and}\ \beta(t,x)=(x-\frac{t}{2})^2-(x+\frac{t}{2})^2

D

α(t,x)=(x−t2)2+2(x+t2)2 and β(t,x)=2(x−t2)2−(x+t2)2\alpha(t,x)=(x-\frac{t}{2})^2+2(x+\frac{t}{2})^2\ \text{and}\ \beta(t,x)=2(x-\frac{t}{2})^2-(x+\frac{t}{2})^2

Answer

α(t,x)=(x−t2)2+(x+t2)2 and β(t,x)=(x−t2)2−(x+t2)2\alpha(t,x)=(x-\frac{t}{2})^2+(x+\frac{t}{2})^2\ \text{and}\ \beta(t,x)=(x-\frac{t}{2})^2-(x+\frac{t}{2})^2

Explanation

Solution

The correct option is (C) : α(t,x)=(x−t2)2+(x+t2)2 and β(t,x)=(x−t2)2−(x+t2)2\alpha(t,x)=(x-\frac{t}{2})^2+(x+\frac{t}{2})^2\ \text{and}\ \beta(t,x)=(x-\frac{t}{2})^2-(x+\frac{t}{2})^2.