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

Answer 1

# y= Ce^x - x #

When we have a First Order Linear non-homogeneous Ordinary Differential Equation of the following form, we can use an integrating factor;

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

We have:

# dy/dx=y+x-1#

# :. dy/dx-y = x-1 # ..... [1]

This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, #I#, using;

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

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

# dy/dxe^( -x )-ye^( -x ) = xe^( -x )-e^( -x ) #

# :. d/dx{ye^( -x )} = xe^( -x )-e^( -x ) #

Our original ODE has now become a Separable ODE as a result of which we can "separate the variables" to obtain::

# ye^( -x ) = int \ xe^( -x )-e^( -x ) \ dx #

When we integrate, we obtain:

# ye^( -x ) = -xe^-x + C #

Finally, arriving at the clear General Solution:

# y= Ce^x - x #

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Answer 2

Isabella Ackerman

The general solution of the given differential equation ( \frac{dy}{dx} = y + x - 1 ) is ( y(x) = Ce^x - x - 2 ), 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.