English: So if you write it as $\frac{d}{dx}\left\lbrack y(x)\ y^{'}(x)\ \right\rbrack = \left\lbrack 0\ 1\ - q(x)\ - p(x)\ \right\rbrack\left\lbrack y(x)\ y^{'}(x)\ \right\rbrack + \left\lbrack 0\ f(x)\ \right\rbrack$ with the initial value being[y(x₀) y^(′)(x₀) ] = [y₀ y₁ ] So you view the second-order system like this.
Telugu: కాబట్టి మీరు దానిని $\frac{d}{dx}\లెఫ్ట్\lbrack y(x)\ y^{'}(x)\ \కుడి\rbrack = \left\lbrack 0\ 1\ - q(x)\ - p(x)\ \కుడి\rbrack\లెఫ్ట్\lbrack y(x)\ y^{'}(x)\ \కుడి\rbrack + \left\lbrack 0\f(x)\ \కుడి\lbrack 0\rbrack 0\rbrack 0\y(x)