Laplace transformation for $T_{1}\cdot V'(t) + V(t) = K\cdot P(t) + Va$

Question

I am trying to do a Laplace transform of the following equation:

$$T_{1}\cdot V'(t) + V(t) = K\cdot P(t) + Va$$

The purpose is to solve $V(s)$ equation knowing that $V(0) = Va$. I followed all the transformation rules and I got the following result:

$$V(s) = \frac{K\cdot P(s)}{s\cdot (1 + T_{1})} + \frac{Va}{s}$$

But, according to the correction, the expected result is :

$$V(s) = \frac{K\cdot P(s)}{(1 + s\cdot T_{1})} + \frac{Va}{s}$$

Did I do something wrong in my calculations?

Answer 1

$$T_{1}\cdot V'(t) + V(t) = K\cdot P(t) + Va$$ Apply Laplace Transform: $$T_{1}(sV(s)-V(0)) + V(s) = K\cdot P(S) + \dfrac {Va}s$$ $$V(s)(sT_{1} + 1) = K\cdot P(S) + \dfrac {Va}s+T_1V(0)$$ Since $V(0)=Va$ we have: $$V(s)(sT_{1} + 1) = K\cdot P(S) + \dfrac {Va}s+T_1Va$$ $$V(s)(sT_{1} + 1) = K\cdot P(S) + \dfrac {Va(1+sT_1)}s$$ It finally gives: $$V(s) = K\cdot \dfrac {P(S)}{(sT_{1} + 1)} +\dfrac {Va}s$$ So yes the result of the correction is correct.