Isa Trevino
2021-01-02
Answered

Solve ${y}^{\u2033}\text{}+\text{}3{y}^{\prime}\text{}-\text{}10y=x({e}^{x}\text{}+\text{}1)$

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SabadisO

Answered 2021-01-03
Author has **108** answers

The auxiliary equation ${m}^{2}\text{}+\text{}3m\text{}-\text{}10=0$

$m=-5,\text{}2$

the complimentary function ${y}_{c}={C}_{1}{e}^{-5x}\text{}+\text{}{C}_{2}{e}^{2x}$

Now The P.I. of given differential equation is

${y}_{p}=\frac{1}{{D}^{2}\text{}+\text{}3D\text{}-\text{}10}x({x}^{x}\text{}+\text{}1)=\frac{1}{(D\text{}+\text{}5)(D\text{}-\text{}2)}x{e}^{x}\text{}+\text{}1$

${e}^{x}\frac{1}{(D\text{}+\text{}1\text{}+\text{}5)(D\text{}+\text{}1\text{}-\text{}2)}x\text{}+\text{}\frac{1}{(D\text{}+\text{}5)(D\text{}-\text{}2)}x$

$={e}^{x}\frac{1}{(D\text{}+\text{}6)(D\text{}-\text{}1)}x\text{}+\text{}\frac{1}{(D\text{}+\text{}5)(D\text{}-\text{}2)}x$

$={e}^{x}\frac{1}{-6(1\text{}+\text{}\frac{D}{6})(1\text{}-\text{}D)}x\text{}+\text{}\frac{1}{-10(1\text{}+\text{}\frac{D}{5})(1\text{}-\text{}\frac{D}{2})}x$

$=\text{}-\frac{{e}^{x}}{6}{(1\text{}+\text{}\frac{D}{6})}^{-1}(1\text{}-\text{}D{)}^{-1}x\text{}-\text{}\frac{1}{10}{(1\text{}-\text{}\frac{D}{5})}^{-1}{(1\text{}-\text{}\frac{D}{2})}^{-1}x$

$=\text{}-\frac{{e}^{x}}{6}(1\text{}+\text{}\frac{D}{6}\text{}+\text{}\cdots )(1\text{}+\text{}D\text{}+\text{}{D}^{2})x\text{}-\text{}\frac{1}{10}(1\text{}+\text{}\frac{D}{5}\text{}\cdots )(1\text{}+\text{}\frac{D}{2}\text{}+\text{}\cdots )x$

Answer

$y={y}_{c}\text{}+\text{}{y}_{p}={C}_{1}{e}^{-5}x\text{}+\text{}{C}_{2}{e}^{2}x\text{}+\text{}-{e}_{6}(x\text{}+\text{}\frac{7}{6})\text{}-\text{}\frac{1}{10}(x\text{}+\text{}\frac{7}{10})$

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