How to Find Two Linearly Independent Solutions of (y' + f(x)y)' = 0?

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The discussion focuses on finding two linearly independent solutions for the differential equation (y' + f(x)y)' = 0, where f is a continuous function on R. The conventional method of differentiating the equation is deemed ineffective due to the lack of differentiability of f. Instead, the approach involves recognizing that the expression in parentheses is a constant, leading to the equation y' + f(x)y = C, where C is an arbitrary constant. By selecting two distinct values for C, such as 0 and 1, one can derive two independent linear first-order equations to solve for the required solutions.

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shapiro478
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Say f is a continuous function on R. How could I find two linearly independent solutions of (y' + f(x)y)' = 0? Notice that there is no hypothesis about f being differentiable, so the obvious method of attack (taking the derivative of each term in the parenthesis and working off the resultant second-order differential equation) probably isn't a good idea. How does the linearly independent part play into this all?
 
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You know the derivative of the left-hand-side is 0, so the bit in parentheses is a constant. That's probably the place to start.
 
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Why in the world would the "obvious method of attack" be to differentiate? As dhris said, the integral will be a constant. y' + f(x)y= C where C is an arbitrary constant. Taking two different values for C, say 0 and 1, will give you two different linear, first order, equations to solve for the two independent solutions to the original equation.
 

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