Find the general solution to the equation displaystyle{x}{left(frac{{left.{d}{y}right.}}{{left.{d}{x}right.}}right)}+{3}{left({y}+{x}^{2}right)}=frac{{ sin{{x}}}}{{x}}

Find the general solution to the equation displaystyle{x}{left(frac{{left.{d}{y}right.}}{{left.{d}{x}right.}}right)}+{3}{left({y}+{x}^{2}right)}=frac{{ sin{{x}}}}{{x}}

Question
Differential equations
asked 2021-02-20
Find the general solution to the equation \(\displaystyle{x}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}+{3}{\left({y}+{x}^{2}\right)}=\frac{{ \sin{{x}}}}{{x}}\)

Answers (1)

2021-02-21
Write this equation as follows:
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{3}\frac{y}{{x}}+\frac{{ \sin{{x}}}}{{{x}^{2}}}-{3}{x}{\left({1}\right)}\)
We first solve the differential equation
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{3}\frac{y}{{x}}={0}\Rightarrow\frac{{\left.{d}{y}\right.}}{{y}}=-{3}\frac{{\left.{d}{x}\right.}}{{x}}\Rightarrow\int\frac{{\left.{d}{y}\right.}}{{y}}=-{3}\int\frac{{\left.{d}{x}\right.}}{{x}}\)
Thus,
\(\displaystyle{\ln}{\left|{y}\right|}=-{3}{\ln}{\left|{x}\right|}+{C}=\frac{ \ln{{1}}}{{{\left|{x}^{3}\right|}}}+{C}\)
Here we used that \(\displaystyle\alpha \ln{\beta}={I}{n}\beta^{\alpha}\). Now use the exponential function:
\(\displaystyle{\left|{y}\right|}=\frac{1}{{{\left|{x}^{3}\right|}}}\cdot{e}^{C}\)
Recall that \(\displaystyle{\left|\alpha\right|}=\pm\alpha\), so we can write
\(\displaystyle{y}=\pm{e}^{C}\cdot\frac{1}{{x}^{3}}\)
Defining a new constant \(\displaystyle{D}=\pm{e}^{C}\) yields
\(\displaystyle{y}=\frac{D}{{x}^{3}}\)
This is the solution of the homogeneous equation. To obtain the solution of the initial equation, we regard D as a function of x, so
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{D}'{x}^{3}-{3}{D}{x}^{2}}}{{{x}^{6}}}=\frac{{{D}'}}{{x}^{3}}-{3}\frac{{{D}}}{{x}^{4}}\)
Plugging this and (2) into (1) yields
\(\displaystyle\frac{{{D}'}}{{x}^{3}}-{3}\frac{D}{{x}^{4}}+{3}\frac{D}{{x}^{4}}=\frac{{ \sin{{x}}}}{{x}^{2}}-{3}{x}\Rightarrow\frac{{{D}'}}{{x}^{3}}=\frac{{ \sin{{x}}}}{{x}^{2}}-{3}{x}\)
Thus,
\(\displaystyle{D}'={x} \sin{{x}}-{3}{x}^{4}\Rightarrow{D}=\int{\left({x} \sin{{x}}-{3}{x}^{4}\right)}{\left.{d}{x}\right.}=\int{x} \sin{{x}}{\left.{d}{x}\right.}-{3}\int{x}^{4}{\left.{d}{x}\right.}\)
Now,
\(\displaystyle\int{x} \sin{{x}}{\left.{d}{x}\right.}={\left\lbrace{\left({u}={x}\Rightarrow{d}{u}={\left.{d}{x}\right.}\right)},{\left({d}{v}= \sin{{x}}{\left.{d}{x}\right.}\Rightarrow{v}=- \cos{{x}}\right)}\right\rbrace}\)
\(\displaystyle=-{x} \cos{{x}}+\int \cos{{x}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{x} \cos{{x}}+ \sin{{x}}+{C}\)
Therefore,
\(\displaystyle{D}=-{x} \cos{{x}}+ \sin{{x}}-\frac{{{3}{x}^{5}}}{{5}}+{C}\)
Finally, plugging this into (2) yields the solution:
\(\displaystyle{y}=-\frac{{ \cos{{x}}}}{{x}^{2}}+\frac{{ \sin{{x}}}}{{x}^{3}}-\frac{3}{{5}}{x}^{2}+\frac{C}{{x}^{3}}\)
0

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