Consider the inhomogeneous partial differential equation in T(x, t) given by ə 1 OT Ox Ox(3+2x²) Choose v(x) such a way that the substitution T(x, t) = v(x) + 6(x, t) gives rise to a homogeneous partial differential equation in p(x, t). 0 ỐT CT = 4 + + ôt ôt² v(x) = 4x² + 4xª O It is not possible to find v(x). v(x) = 4x²8x¹ B v(x)=8x² + 4x v(x) 6x²+2x+

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