Find the general solution of the differential equation? y dy/dx-2e^x=0

1 Answer
Apr 15, 2018

#y = pm 2sqrt(e^x + A)#

Explanation:

We wish to solve the differential equation #y * y' - 2e^x = 0#.

This is a separable, first-order ordinary differential equation. As such, it can be solved using techniques suitable for separable 1st-order ODE's.

The most straightforward technique is to get our equation in the form #f(y) dy = g(x) dx#. We can then integrate both sides, ridding ourselves of the #y'# term.

We get our function into this form as such:

#y dy/dx - 2e^x = 0#
#y dy/dx = 2e^x#
#(y) dy = (2e^x) dx#

See that our left-hand side is a function of just #y# and our right-hand side a function of just #x#. Integrate both sides.

#int (y) dy = int (2e^x)dx#
#1/2 y^2 = 2e^x + C#
#y^2 = 4e^x + 2C#
#y = pm sqrt(4e^x + 2C) = pm 2sqrt(e^x + C/2)#

Since #C# is an arbitrary constant, let #A = C/2#. Then our final answer is #y = pm 2sqrt(e^x + A)#.