Solve the linear equations by considering y as a function of x, that is y=y(x) displaystyle{y}'-{2}{x}{y}={e}^{{{x}^{2}}}

geduiwelh 2020-12-28 Answered
Solve the linear equations by considering y as a function of x, that is
y=y(x)
y2xy=ex2
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Expert Answer

brawnyN
Answered 2020-12-29 Author has 91 answers

Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
y=2xydydx=2xydyy=2xdx
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
dyy=2xdx
Which is
ln|y|=x2+c
By taking exponents, we obtain
|y|=ex2+c=ex2ec
Hence,we obtain
y=Cex2
where C=±ecandyc=ex2 is the complementary solution .
Next, we need to find the particular solution yp
Therefore, we consider uyc, and try to find u, a function of x, that will make this work.
Let's assume that uyc is a solution of the given equation. Hence, it satisfies the given equation.
Substituting uyc, and its derivative in the equation gives
(uyc)+2xuyc=ex2
uyc+uyc+2xuyc=ex2
uyc+u(yc+2xyc)=0  =ex2
since yc is a solution
Therefore,
uyc=ex2u=ex2yc
which gives
u=ex2ycdx
Now, we can find the function u:
u=(ex2ex2)dx
=dx
=x+c
Since we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,
u=x
Recall that yp=uyc
Therefore
yp=xex2
The general solution is
y=Cyc+yp
=Cex2+xex2
=ex2(x+C)
Integration Factor technique
This equation is linear with P(x)=2xandQ(x)=ex2
Hence,
h=

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