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For $v=(t+1{)}^{2}u$, the derivatives areSubstitution cancels the two terms $2(t+1)u$, giving the diffusion equationThis is a scalar integrating factor for a reaction-diffusion equation, since $\exp [\int 2dt/(t+1)]=(t+1{)}^2$ up to a constant. We use $t\ge 0$, so the factor never vanishes. A classical solution at the initial corners additionally requires compatible data, $h(0)=0$ and $h(\pi )=f(0)$.