English: Then the initial value problem y^(″) + p(x) y^(′) + q(x)y = f(x), along with initial conditions y(x₀) = y₀, and y′(x₀) = y₁ has a unique So the proof is beyond the scope of this course, what we do is, we will just give the justification why this theorem is true.
Kannada: ನಂತರ ಆರಂಭಿಕ ಮೌಲ್ಯದ ಸಮಸ್ಯೆ y^(″) + p(x) y^(′) + q(x)y = f(x), ಆರಂಭಿಕ ಷರತ್ತುಗಳೊಂದಿಗೆ y(x₀) = y₀, ಮತ್ತು y′(x₀) = y₁ ವಿಶಿಷ್ಟ ಪರಿಹಾರ y(x) ∀x ∈ I ಹೊಂದಿದೆ.