Question #5a53b

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Answer 1 · 2017-03-24T00:16:04

Looks like you're nailing it. Here's how I would have a crack at this.

#xy + y' = 100x#

If we play with the algebra first:

# y'= x (100 - y)#

This is separable:

# (y')/ (100 - y) = x#

So we integrate both sides wrt x:

#int 1/ (100 - y) \ dy/dx \dx = int x \ dx#

By Chain Rule:
#implies int 1/ (100 - y) \dy = int x \ dx#

And so:

# -ln (100 - y) =x^2/2 + C#

# ln (100 - y)^(-1) =x^2/2 + C#

Take each term as an exponential:

# e^( ln (100 - y)^(-1) ) = e^ ( x^2/2 + C) #

# (100 - y)^(-1) = e^C e^ ( x^2/2) = C e^ ( x^2/2)#

# 100 - y = C e^ ( -x^2/2) #

#y = 100 - C e^ ( -x^2/2) #