Solve the linear equations by considering y as a function of x, that is, y = y(x). displaystylefrac{{{left.{d}{y}right.}}}{{{left.{d}{x}right.}}}-{y}={4}{e}^{x},{y}{left({0}right)}={4}

shadsiei 2021-02-25 Answered
Solve the linear equations by considering y as a function of x, that is, y = y(x).
dydxy=4ex,y(0)=4
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Expert Answer

2abehn
Answered 2021-02-26 Author has 88 answers

Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
dydx=ydyy=dx
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=dx
which isln|y|=x+c
By taking exponents, we obtain
|y|=ex+c=exec
Hence,we obtain
y=Cex
where C=±ecandyc=ex is the compleentary 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)uyc=4ex
uyc+uycuyc=4ex
uyc+u(ycyc)=0  =4ex
since yc is a solution
Therefore
uyc=4exu=4exyc
which gives
u=exycdx
Now, we can find the function u :
u=4exexdx
=4dx
=4x+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=4x
Recall that yp=uyc. Therefore,
yp=4xex
The general solution is
y=Cyc+yp
=ex(C+4x)
Integrating Factor technique
This equation is linear with P(x)=1andQ(x)=4ex
Hence,
h=Pdx=dx=x
So, an integrating factor is
eh=ex
and the general solution is
y(x)=eh(c+Qehdx)
=ex(c+4exex
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