# Finding value of a constant in Differential Equations

I have the adhering to ODE Where offered is $x(0)=1$:

$$(t+3)dx=4x^2dt$$

After splitting up of variables I obtained this:

$$\frac{-1}{x} = 4\ln(t+3)+C$$

I assume this streamlines extra as:

$$x=\frac{-1}{\ln((t+3)^4)+C}$$

Please inform me if this is proper, I additionally have trouble searching for C in this instance, MapleTA does decline my solution which takes place to be:

$$x=\frac{-1}{\ln((t+3)^4)+(-0.52)}$$

**Update: **

I located that the C needs to e expresed additionally in $\ln$ (logarithmic) term, not a number. Any kind of suggestions?

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Sean87 2022-07-25 20:40:38

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