Solve the linear equations by considering y as a function of x, that is, \displaystyle{y}={y}{\left({x}\right)}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}-{2}\frac{x}{{{1}+{x}^{2}}}{y}={x}^{2}

Ava-May Nelson 2020-11-20 Answered

Solve the linear equations by considering y as a function of x, that is, y=y(x).dydx2x1+x2y=x2

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Expert Answer

Luvottoq
Answered 2020-11-21 Author has 95 answers

Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
y=y(x).dydx2x1+x2y=x2
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=2xdx1+x2
Let’s solve the integral on the right side.
2xdx1+x2=|1+x2=t2xdx=dt|
=dtt
=ln|t|+x
=ln|1+x2|+c1+x2>0xR
Therefore
In|y|=ln(1+x2)+c.
By taking exponents, we obtain
ln|y|=ln1+x2+c=(1+x2)ec
Hence,we obtain
y=C(1+x2)
where C=±ecandy,=1+x2 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)uyc2x1+x2=x2
uyc+uyc2x1+x2=x2
uyc+u(ycoprace2x1+x2)=0 since yc is a solution=x2
Therefore,
uyc=x2u=x2yc
which gives
u=x2ycdx
Now, we can find the function u:
u=x2+111+x2
=(111+x2
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