BenoguigoliB

2021-03-09

Solve differential equation $\frac{dy}{dx}+\left(\frac{a}{x}\right)y=40x$, for $x>0$ and $y\left(1\right)=a$

ottcomn

$\frac{dy}{dx}+P\left(x\right)×y=Q\left(x\right)$
$I.F.=e\left(\int Pdx\right)$
$y\left(I.F.\right)=\int Qx\left(I.F.\right)dx+C$
$P\left(x\right)=\frac{a}{x}$
$Q\left(x\right)=40x$
$I.F.=e\left(\int \frac{a}{x}dx\right)={e}^{a\mathrm{ln}x}={e}^{\mathrm{ln}{x}^{a}}={x}^{a}$
$y\left(I.F.\right)=\int \left(40x\right)\left(I.F.\right)dx+C$
$y\left({x}^{a}\right)=\int \left(40x\right)\left({x}^{a}\right)dx+C$
$y{x}^{a}=40\int {x}^{a+1}dx+C$
$y{x}^{a}=40×\frac{{x}^{2}+1+1}{a+1+1}+C$
$y{x}^{a}=40×\frac{{x}^{a+2}}{a+2}+C$
$y=\frac{40}{a+2}×\frac{{x}^{a+2}}{{x}^{a}}+\frac{C}{{x}^{a}}$
$y=\frac{40}{a+2}×{x}^{2}+\frac{C}{{x}^{a}}$ (1)
$y\left(1\right)=a$
Put $x=1$ and $y=a$ into equation (1)
$a=\frac{40}{a+2}\left(1{\right)}^{2}+\frac{C}{{1}^{a}}$
$a=\frac{40}{a+2}+C$
$C=a-\frac{40}{a+2}$
$C=\frac{a\left(a+2\right)-40}{a+2}$
$C=\frac{{a}^{2}+2a-40}{a+2}$ Put the value of C into equation (1) $y=\frac{40}{a+2}{x}^{2}+\frac{{a}^{2}+2a-40}{a+2}×\frac{1}{{x}^{a}}$

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