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\nSolving this differential equation, I am stuck at the initial condition
\nLet the equation be this:\n$$y''-\\frac{2y'}{y}=2$$\nAnd the initial conditions are\n$$y(0)=\\alpha$$\n$$y'(0)=0$$\nSolving this equation is easy, since $\\alpha$ is arbitrary, then $y(t)=\\alpha t+\\beta$ and the problem then just becomes\n$$\\beta=\\alpha\\alpha^2+\\alpha-2$$\nIs this correct?
\nA:
\nWhen you set $y(0)=\\alpha$, you're given an additional requirement over and above the ODE and the two initial conditions. You don't need to solve the DE, instead, once you integrate you get that\n$$\ny(t)=\\alpha\\cdot t+c\n$$\nFor reasons of format, I'll put this in $\\alpha$ so that we can work with $\\alpha^2$:\n$$\ny(t)=\\alpha\\cdot t+c\n$$\nwhere $\\alpha$ and $c$ are both known. We want to find $y'(t)$ so we differentiate w.r.t. $t$:\n$$\n\\frac{y'(t)}{y(t)}=\\frac{\\alpha+c'}{\\alpha\\cdot t+c}=\\frac{\\alpha}{\\alpha\\cdot t+c}\n$$\nThat gives $y'(t)=\\frac{\\alpha}{(\\alpha\\cdot t+c)^2}$ so we can set the initial conditions and solve for $t$:\n$$\ny'(0)=\\frac{\\alpha}{\\alpha+c}=0\n$$\n$$\n\\alpha\\cdot t+c=0\n$$\nhttps://documenter.getpostman.com/view/21869250/UzkdtrR5 https://documenter.getpostman.com/view/21846092/UzkdtrR6 https://documenter.getpostman.com/view/21838696/UzkdtrLn https://documenter.getpostman.com/view/21842012/UzkdtrLm https://documenter.getpostman.com/view/21854109/UzkdtrR7 # Introduction\nWhat does your API do?
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