y'' + 2x (y')^2= 0 show the general solution, then find an explicit formula for a solution y(x), which also satisfies intial conditions y'(0) =1 and y (0) = 0 where to start?
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Originally Posted by harveyo y'' + 2x (y')^2= 0 show the general solution, then find an explicit formula for a solution y(x), which also satisfies intial conditions y'(0) =1 and y (0) = 0 where to start? Let . The equation becomes . Apply separation of variables to solve for u, where . Then back substitute and solve the resulting differential equation for y. Can you take it from here?
seperation of variables = u= -2x u^2 dy/dx = -2x u^2 dy/u^2 = -2x dx integrate both sides ln u^2 = -x^2 + C am i doing this correct
Originally Posted by harveyo seperation of variables = u= -2x u^2 dy/dx = -2x u^2 dy/u^2 = -2x dx integrate both sides ln u^2 = -x^2 + C am i doing this correct no. (!) Note that by the way, why do you have dy? it should be du
du/dx = -2x u^2 du/u = -2x dx intergrate u ^-2 du = intergrate -2x dx -u ^-1 = -x^+c
Originally Posted by harveyo du/dx = -2x u^2 du/u = -2x dx You mean du/u^2 intergrate u ^-2 du = intergrate -2x dx -u ^-1 = -x^+c Good. so And now, since , you still need to solve That should be an easy integral and will introduce a second "constant of integration".
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