Solve the linear equations by considering y as a function of x, that is, y=y(x).xy′+(1+x)y=e−xsin⁡2x

Carol Gates

Carol Gates

Answered

2020-10-28

Solve the linear equations by considering y as a function of x, that is, y=y(x).xy+(1+x)y=exsin2x

Answer & Explanation

Elberte

Elberte

Expert

2020-10-29Added 95 answers

Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
xy=(1+x)ydydx=(1+x)yxdy=(1+x)dxx
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
xy=(1+x)ydydx=(1+x)yxdy=(1+x)dxx
Let’s solve the integral on the right side.
(1+x)dxx=(1x+xx)dx
=(1x+1)dx
=dxx+dx
=ln|x|+x+c
Therefore,
ln|y|=ln|x|x+c
By taking exponents, we obtain
|y|=eln|x|x+c=1|x|exec
Hence,we obtain
y=Cxex
where C=±ecandyc=1xex 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.
Write the equation in the standard form (divide it by x 4 0)
y+(1+x)yx=exsin2xx
Substituting displaysty{u}{y}{c} and its derivative in the equation gives
uyc
(uyc)+(1+x)uycx=exsin2xx
uyc+u(yc+(1+x)ycx)=0 since ycis a solution

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