derivative review: find y' for y=4^x^3, solve xy' - 3y = 0
- Thread starter paulxzt
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skeeter
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\(\displaystyle \L y = 4^u\)
\(\displaystyle \L y' = 4^u \cdot \ln{4} \cdot \frac{du}{dx}\)
your u is x<sup>3</sup>.
the second problem is a differential equation ... solve for y
\(\displaystyle \L xy' - 3y = 0\)
\(\displaystyle \L xy' = 3y\)
\(\displaystyle \L y' = \frac{3y}{x}\)
\(\displaystyle \L \frac{dy}{dx} = \frac{3y}{x}\)
\(\displaystyle \L \frac{1}{y} dy = \frac{3}{x} dx\)
\(\displaystyle \L \ln|y| = 3\ln|x| + C_1\)
\(\displaystyle \L y = C_2 x^3\)
check ... \(\displaystyle \L y' = 3C_2 x^2\)
\(\displaystyle \L xy' - 3y = x \cdot 3C_2 x^2 - 3 \cdot C_2 x^3 = 0\) ... checks good.
\(\displaystyle \L y' = 4^u \cdot \ln{4} \cdot \frac{du}{dx}\)
your u is x<sup>3</sup>.
the second problem is a differential equation ... solve for y
\(\displaystyle \L xy' - 3y = 0\)
\(\displaystyle \L xy' = 3y\)
\(\displaystyle \L y' = \frac{3y}{x}\)
\(\displaystyle \L \frac{dy}{dx} = \frac{3y}{x}\)
\(\displaystyle \L \frac{1}{y} dy = \frac{3}{x} dx\)
\(\displaystyle \L \ln|y| = 3\ln|x| + C_1\)
\(\displaystyle \L y = C_2 x^3\)
check ... \(\displaystyle \L y' = 3C_2 x^2\)
\(\displaystyle \L xy' - 3y = x \cdot 3C_2 x^2 - 3 \cdot C_2 x^3 = 0\) ... checks good.