What is the general solution of the differential equation # dy/dx- 2xy = x #?

Answer 1

# y = 3/2e^(x^2 - 1) - 1/2 #

We have:

# dy/dx- 2xy = x #

Which we can write as:

# dy/dx = 2xy+ x # # :. dy/dx = (2y+ 1)x # # :. 1/(2y+ 1) dy/dx = x #

Which is a first order separable differential equation, so we can "separate the variables" to get:

# int \ 1/(2y+ 1) \ dy = int \ x \ dx #

Integrating we get, the General Solution:

# 1/2ln|2y+1| = 1/2x^2 + C #
Applying the initial condition #y(1)=1# we find:
# 1/2ln3 = 1/2 + C => C = 1/2ln3 - 1/2#

So we can write an implicit particular solution as:

# 1/2ln|2y+1| = 1/2x^2 + 1/2ln3 - 1/2 #

We typically require an explicit solution, so we can rearrange as follows:

# ln|2y+1| = x^2 + ln3 - 1 # # :. |2y+1| = e^(x^2 + ln3 - 1) #
Noting that the exponential function is positive over its entire domain, (as #e^x gt 0 AA x in RR#):
# 2y+1 = e^(x^2 - 1)e^(ln3) #
# :. 2y = 3e^(x^2 - 1) - 1 #
# :. y = 3/2e^(x^2 - 1) - 1/2 #
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Answer 2

The general solution of the differential equation ( \frac{dy}{dx} - 2xy = x ) is given by ( y(x) = Ce^{x^2} + \frac{1}{2}x - \frac{1}{4} ), where ( C ) is the constant of integration.

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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.

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.

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.

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.

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