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Solution to Differential Equation Passing Through (0,e)
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[QUOTE="mrchauncey, post: 4662954, member: 502571"] [h2]Homework Statement [/h2] Find the solution to the differential equation. Which passes through the point (0,e). [itex]\frac{dy}{dx}[/itex] = [itex]\frac{11xy}{(ln(y))^{10}}[/itex] [h2]Homework Equations[/h2] I can get through the integration part but I am lost when it comes to using the ln rules to find the solution. [h2]The Attempt at a Solution[/h2] First step I move the variables to each side. [itex]\frac{(ln(y))^{10}}{y}[/itex] dy = 11x dx I then integrate each side: ∫[itex]\frac{(ln(y))^{10}}{y}[/itex] dy = ∫11x dx ∫[itex]\frac{(ln(y))^{10}}{y}[/itex] dy, I used sub method. u = ln(y), du = 1/y dy -> dy = ydu ∫[itex]\frac{u^{10}}{y}[/itex] * y du, so the y's cancel out. making it: ∫ u[itex]^{10}[/itex] du = [itex]\frac{u^{11}}{11}[/itex] = [itex]\frac{ln(y)^{11}}{11}[/itex] The other side is simple integration: ∫11x dx = [itex]\frac{11x^{2}}{2}[/itex], this gives: [itex]\frac{ln(y)^{11}}{11}[/itex] = [itex]\frac{11x^{2}}{2}[/itex] + c, since it passes through the point (0,e), I find my constant value c right away. y(0)=e =[itex]\frac{ln(e)^{11}}{11}[/itex] = [itex]\frac{11(0)^{2}}{2}[/itex] + c y(0)=e = 1 = 0 + c, so c = 1. I then multiplied the 11 by both sides and got: (ln(y))[itex]^{11}[/itex] = [itex]\frac{121x^{2}}{2}[/itex] + 11 This is where I am getting confused. I do not know how to use the ln rules on this one. The power of 11 is confusing me and I can not find an example like this one. Any help would be nice. thanks. [/QUOTE]
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Solution to Differential Equation Passing Through (0,e)
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