Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this?
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Hello Originally Posted by jaijay32 Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this? Not sure if I get that right You want to solve ? Using the product rule leads to ( u' = 1/x and v = ln(x) ) Add dx to both sides devide by 2 Was it your question? Yours Rapha
Originally Posted by jaijay32 Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this? dy/dx = y{[ln x]/x} Rearranging, [1/y][dy]=[ln x]/[x] dx integrate both sides, ln y = integrate{[lnx]/x} dx ln y = 0.5(ln x)^2 +c Notice when you differentiate 0.5(ln x)^2 with respect to x you'll get back [ln x]/x. I hope I helped.
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