Differentiate the e e

Calculus Level 1

Differentiate : d y d x = e 3 x + 2 y \displaystyle{\frac{dy}{dx} = e^{3x+2y}}

y = − 1 2 ln ⁡ ( − 2 ( 1 3 e 3 x + K ) ) y= -\frac{1}{2} \ln(-2(\frac{1}{3}e^{3x}+K)) y = − 1 2 ln ⁡ ( − 5 ( 1 3 e 3 x + K ) ) y= -\frac{1}{2} \ln(-5(\frac{1}{3}e^{3x}+K)) y = − 1 2 ln ⁡ ( 2 ( 1 4 e 3 x + K ) ) y= -\frac{1}{2} \ln(2(\frac{1}{4}e^{3x}+K)) y = − 1 4 ln ⁡ ( − 2 ( 1 3 e 3 x + K ) ) y= -\frac{1}{4} \ln(-2(\frac{1}{3}e^{3x}+K))

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1 solution

Follow the rules of the first order separable ODE: N ( y ) d y = M ( x ) d x N(y)dy = M(x) dx to get the answer.

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How to use this rule.. Would you please elaborate it..

Ravi Hammad - 5 years, 2 months ago

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Go to www2.fiu.edu read how to elaborate then to solve.

A Former Brilliant Member - 5 years, 2 months ago

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ADIOS YO YO HONEY SINGH.

Manish Maharaj - 5 years, 1 month ago

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@Manish Maharaj – ⌣ ¨ \ddot \smile

A Former Brilliant Member - 5 years, 1 month ago

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