1 answer

How can i solve this equation ?

Question:

y'-y=e^2x using linear defferential equation


Answers

#y = e^x (e^x + C)#

Explanation:

Simplest approach is to use an Integrating factor.

For:

  • #y' + f(x) y = g(x)#,

there may be a function:

  • #bb I(x) = exp (int f(x) dx)#,

which when applied to the LHS makes it exact.

Ie if we can find #bb I (x)#, then:

  • #bb I(x) y' + bb I(x) f(x) y = (bb I(x) y)^'#

So here #bb I = e^(int (-1) dx) = e^(-x + C) = Ce^(-x)#

It is usual to drop the constant #C #, as it cancels out on both sides, so we have:

#y' - y = e^(2x)#

#e^(-x)y' - e^(-x)y = e^(-x)* e^(2x)#

#(e^(-x)y)^' = e^(x)#

#e^(-x)y = e^(x) + C#

#y = e^(2x) + Ce^x = e^x (e^x + C)#

.

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