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differential equation from calculus II

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Given the differential equation: (y^4)(e^2x) + y' = 0

NOTE: The differential equation above is attached in a microsoft word document for better legibility. Additionally my work is attached as a jpeg file.

The questions:

a)Find the general solution.

b)Find the particular solution such that y(0) = 1.

https://brainmass.com/math/calculus-and-analysis/differential-equation-calculus-4616

SOLUTION This solution is FREE courtesy of BrainMass!

y^4*e^2x +dy/dx = 0

or dy/dx = -y^4*e^2x

or (-1/y^4)dy = e^2x dx

Integrating we have 3/y^3 = 1/2 e^2x + C

or 3/(1/2 e^2x +c ) = y^3...General Solution

Putting x = 0 we have 3/( 1/2 + c ) = 1

or 1/2 + c = 3

so c = 5/2

So particular solution = 3/(1/2 e^2x +5/2 ) = y^3

Hope it is clear

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!