Let's start with the given differential equation: \[ p(x)x''(t) + q(t)x'(t) + r(t)x(t) = 0 \] We are to show that the equation has the same solutions as: \[ (a(t)x'(t))' + b(t)x(t) = 0, \] where \( a'(x) \) is continuous and \( b(x) \) is continuous. --- ## Step 1: **Rewrite the Original Equation** Start with: \[ p(t)x''(t) + q(t)x'(t) + r(t)x(t) = 0 \] --- ## Step 2: **Express as a Derivative** We want to write the first two terms as the derivative of a product: \[ \frac{d}{dt}(a(t)x'(t)) = a'(t)x'(t) + a(t)x''(t) \] Compare this with the original: \[ p(t)x''(t) + q(t)x'(t) \] Set the coefficients equal: \[ a(t)x''(t) + a'(t)x'(t) \equiv p(t)x''(t) + q(t)x'(t) \] So, **matching coefficients:** - \( a(t) = p(t) \) - \( a'(t) = q(t) \) --- ## Step 3: **Solve for \( a(t) \) in Terms of \( p(t) \) and \( q(t) \)** From above: \[ a(t) = p(t) \] \[ a'(t) = q(t) \] But with \( a'(t) = q(t) \), we can check consistency with \( p(t) \): \[ a'(t) = \frac{d}{dt}a(t) = \frac{d}{dt}p(t) = p'(t) \] So, to match \( a'(t)x'(t) = q(t)x'(t) \), we need: \[ q(t) = p'(t) \] But in general, \( q(t) \) is not necessarily \( p'(t) \). So, to match the original equation, we need to find an **integrating factor**. --- ## Step 4: **Find Integrating Factor** Divide the whole equation by \( p(t) \) (assuming \( p(t) \neq 0 \)): \[ x''(t) + \frac{q(t)}{p(t)} x'(t) + \frac{r(t)}{p(t)} x(t) = 0 \] Let’s look for an integrating factor \( \mu(t) \) such that: \[ \mu(t) x''(t) + \mu(t) \frac{q(t)}{p(t)} x'(t) + \mu(t) \frac{r(t)}{p(t)} x(t) = 0 \] We want to express the first two terms as a single derivative: \[ \frac{d}{dt}\left[ \mu(t) x'(t) \right] = \mu(t)x''(t) + \mu'(t)x'(t) \] So, **to match coefficients:** \[ \mu'(t) = \mu(t) \frac{q(t)}{p(t)} \] This is a first-order linear ODE for \( \mu(t) \): \[ \frac{\mu'(t)}{\mu(t)} = \frac{q(t)}{p(t)} \] \[ \ln \mu(t) = \int \frac{q(t)}{p(t)} dt \] \[ \mu(t) = \exp\left( \int \frac{q(t)}{p(t)} dt \right) \] --- ## Step 5: **Rewrite the Equation** Now, multiply the original equation by \( \mu(t) \): \[ \mu(t)p(t)x''(t) + \mu(t)q(t)x'(t) + \mu(t)r(t)x(t) = 0 \] But \( \mu(t)p(t)x''(t) + \mu(t)q(t)x'(t) = \frac{d}{dt}\left[ \mu(t)p(t) x'(t) \right] \). So, \[ \frac{d}{dt} \left[ \mu(t) p(t) x'(t) \right] + \mu(t) r(t) x(t) = 0 \] Let \( a(t) = \mu(t) p(t) \), \( b(t) = \mu(t) r(t) \). Both are continuous functions (since \( \mu(t) \), \( p(t) \), and \( r(t) \) are continuous). --- ## **Final Answer** The original equation \[ p(t)x''(t) + q(t)x'(t) + r(t)x(t) = 0 \] has the same solutions as \[ \left[ a(t)x'(t) \right]' + b(t)x(t) = 0 \] where \[ a(t) = p(t) \exp \left( \int \frac{q(t)}{p(t)} dt \right) \] \[ b(t) = r(t) \exp \left( \int \frac{q(t)}{p(t)} dt \right) \] and both \( a(t) \) and \( b(t) \) are continuous. --- ### **Summary** - Find integrating factor \( \mu(t) = \exp(\int \frac{q(t)}{p(t)} dt) \). - Multiply original equation by \( \mu(t) \). - Combine first two terms into a single derivative: \( (a(t)x'(t))' \). - The resulting equation is of the required form.