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I was trying to solve this but I can't not fix the problem :( please help me.

a,b,f and d positive vale.

Clear["Global`*"]
eqns = a*y'[t] == b*y[t]^2 + f*y[t]+d;
DSolve[{eqns, y[0] == 0, y[t0] == 1}, y[t], t]
y[t] = x'[t]/x[t];
DSolve[{eqns, x[t0] = x0}, x[t], t]
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    $\begingroup$ First DSolve : You use 2 initial conditions, one is enough for an ode of first order. t0 isn't defined! Is y[t] = x'[t]/x[t]intended to transform the ode? $\endgroup$ Jan 14 at 14:53
  • $\begingroup$ Okay how can put two initial conditions? t0 is positive and I just want the solution with t0. I would like to use the soluion of y[t] from first DSolve and put to another Dsolve $\endgroup$ Jan 14 at 14:59
  • $\begingroup$ That means: Your 2nd ode is y[t]==x'[t]/x[t]? $\endgroup$ Jan 14 at 15:03
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If I understand your problem correctly, you might solve the problem in one step

sol=DSolve[{a*y'[t] == b*y[t]^2 + f*y[t]+ d, 
y[t] == x'[t]/x[t],
y[t0] == 1}, {x, y}, t]
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  • $\begingroup$ exactly this is what I mean but other problem is come up is : Cos[(Sqrt[4 b d - f^2] (t + (-Sqrt[4 b d - f^2] t0 + 2 a ArcTan[(2 b Sqrt[4 b d - f^2] + f Sqrt[4 b d - f^2])/(4 b d - f^2)])/Sqrt[4 b d - f^2]))/(2 a)]^(-(a/b)) this part should be (Sinh [b/a..t] /Sinh[b/a...t0] )^(-a/b) $\endgroup$ Jan 14 at 15:11
  • $\begingroup$ Simplify[ DSolve[{ay'[t] == by[t]^2 + f*y[t] + d, y[t] == x'[t]/x[t], y[t0] == 1}, {x, y}, t], Assumptions -> {t0 > 0, a > 0, b > 0, f > 0, d > 0}] $\endgroup$ Jan 14 at 15:16
  • $\begingroup$ This is the solution Exp[-f/2b (t-t0)]*(Sinh [-b/a..t] /Sinh[-b/a...t0] )^(-a/b) $\endgroup$ Jan 14 at 15:20
  • $\begingroup$ @FelipeDura Your comments are difficult to understand if you do not tell how you got your results! $\endgroup$ Jan 14 at 15:23
  • $\begingroup$ Can you see the solution now? $\endgroup$ Jan 14 at 15:42

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