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What is the general solution of the differential equation # (1+x)dy/dx-y=1+x #?

Answer 1

Isaiah Allen

# y = (1+x)ln(1+x) + C(1+x) #

We have:

# (1+x)dy/dx-y=1+x #

This ODE can be rearranged as follows:

# dy/dx - 1/(1+x) y=1 # ..... [1]

The following is an example of a First Order Linear non-homogeneous Ordinary Differential Equation:

# dy/dx + P(x)y=Q(x) #

This form of equation, which we can easily create an integrating factor for, is given by;

# I = e^(int P(x) dx) # # \ \ = exp(int \ -1/(1+x) \ dx) # # \ \ = exp( -ln(1+x) ) # # \ \ = -(1+x) #

And if we multiply the DE [1] by this Integrating Factor, #I#, we will have a perfect product differential;

# -1/(1+x) dy/dx + 1/(1+x)^2 y = -1/(1+x) # # :. d/dx[ -1/(1+x) y] = -1/(1+x) #

which we can integrate directly to obtain:

# 1/(1+x) y = int \ 1/(1+x) \ dx # # :. 1/(1+x) y = ln(1+x) + C #

# :. y = (1+x)ln(1+x) + C(1+x) #

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Answer 2

Christopher Allers

The general solution of the given differential equation (1+x)dy/dx - y = 1+x is y = x + C*(1+x), where C is an arbitrary constant.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.