Math Challenges

Submissions for Problem #26

Find the general solution to the given differential equation

\[ \frac{dy}{dt}=\frac{\left(4y+3\right)^3}{t^{\frac12}} \]

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Solution: \( \displaylines{\frac{dy}{\left(4y+3\right)^3}=t^{-\frac12}dt\Rightarrow-\frac18\left(4y+3\right)^{-2}=2t^{\frac12}+C\\ \left(4y+3\right)^2=\frac{1}{-16\sqrt{t}+C}\\ y=\frac14\sqrt{\frac{1}{-16\sqrt{t}+C}}-\frac34} \)