hexacordoK

2020-11-23

Solve differential equation $6\left(y-{x}^{2}\right)dx+xdy=0$

firmablogF

$6\left(y-{x}^{2}\right)dx+xdy=0$
$\left(6\left(y-{x}^{2}\right)dx\right)/\left(xdx\right)+\left(xdy\right)/\left(xdx\right)=0$
$\left(6\left(y-{x}^{2}\right)dx\right)/\left(xdx\right)+\left(xdy\right)/\left(xdx\right)=0$
$6\left(\left(y-{x}^{2}\right)/x\right)+dy/dx=0$ $dy/dx+\left(6/x\right)y=6x$ (1)
Now, equation (1) is first order linear differential equation
We know that solution of first order linear differential equation is given by
$y.\left(I.F.\right)=\int \left(I.F.\right)6xdx$ (2)x
where, $I.F.={e}^{\int 6/xdx}$
$I.F.={e}^{6\int dx/x}$ $\left({:}^{\prime }\int dx/x=\mathrm{ln}x+c\right)$
$={e}^{6\left(\mathrm{ln}x\right)}$
$={e}^{\mathrm{ln}{x}^{6}}$
$={x}^{6}$
Now, solving equation(2) by using I.F.
$y\left({x}^{6}\right)=\int {x}^{6}\left(6x\right)dx$
$=6\int {x}^{7}dx$ $\left({:}^{\prime }\int {x}^{n}dx=\left({x}^{n+1}\right)/\left(n+1\right)+c$
$=6\left({x}^{8}/8\right)+c$
$=\left(3{x}^{8}\right)/4+c$ (3)
Now, using y(1)= 1 in equation (3)
$1\left(1{\right)}^{6}=\left(3\left(1{\right)}^{8}\right)/4+c$
$1=3/4+c$
$c=1-3/4$
$c=\left(4-3\right)/4$
$c=1/4$
Now, putting value of c in equation (3) $y{x}^{6}=\left(3{x}^{8}\right)/4+1/4$
Hence, solution of given differential equation is $y{x}^{6}=\left(3{x}^{8}\right)/4+1/4$

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