Chapter 38: Problem 25
$$ \frac{\mathrm{d} y}{\mathrm{~d} x}+y \tan x=y^{3} \sec ^{4} x $$
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Chapter 38: Problem 25
$$ \frac{\mathrm{d} y}{\mathrm{~d} x}+y \tan x=y^{3} \sec ^{4} x $$
These are the key concepts you need to understand to accurately answer the question.
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$$ \left(x y+y^{2}\right)+\left(x^{2}-x y\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 $$
The rate of decay of a radioactive substance is proportional to the amount A remaining at any instant. If \(A=A_{0}\) at \(t=0\), prove that, if the time taken for the amount of the substance to become \(\frac{1}{2} A_{0}\) is \(T\), then \(A=A_{0} e^{-(t \ln 2) / T}\). Prove also that the time taken for the amount remaining to be reduced to \(\frac{1}{20} A_{0}\) is \(4 \cdot 32 T\).
Solve the following differential equations: $$ (1+x)^{2} \frac{\mathrm{d} y}{\mathrm{~d} x}=1+y^{2} $$
$$ \frac{\mathrm{d} y}{\mathrm{~d} x}+y \tan x=\sin x $$
$$ \frac{\mathrm{d} y}{\mathrm{~d} x}+y=y^{4} e^{x} $$
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