Taking Kolmogorov's forward DE as presented in contingencies, we can simplify the RHS as follows:

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k \neq j} (_tp_x^{ik} \mu_{x+t}^{kj}) - \sum_{k \neq j} (_tp_x^{ij} \mu_{x+t}^{jk})$$

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k \neq j} (_tp_x^{ik} \mu_{x+t}^{kj}) - _tp_x^{ij} \sum_{k \neq j} \mu_{x+t}^{jk}$$

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k \neq j} (_tp_x^{ik} \mu_{x+t}^{kj}) + _tp_x^{ij}\mu_{x+t}^{jj}$$ since \( \mu_{x+t}^{jj} = - \sum_{k \neq j} \mu_{x+t}^{jk} \). Therefore:

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k} (_tp_x^{ik} \mu_{x+t}^{kj}) = \frac{d}{dt} \ _tp_x^{ij} $$

I will leave it to you to verify the DE as derived in your statistics classes as see why the last line above should hold in general.