Hello ,please someone should help with out with these Diffrential equations a. xdy + ydx = x^3 y^6 dx b. cosysin2xdx +(cos^2y - cos^x)dy = 0
Part (a) xdy + ydx = x^3 y^6 dx $x \, dy + y \, dx = x^3 y^6 \, dx$
$dy + \dfrac{1}{x}y \, dx = x^2 y^6 \, dx$
$Q = x^2$
$n = 6$
$(1 - n) = -5$
$z = y^{1 - n} = y^{-5}$
Integrating Factor $\begin{align} \displaystyle u & = e^{(1 - n)\int P\,dx} \\ & = e^{-5\int \frac{1}{x}\,dx} \\ & = e^{-5 \ln x} \\ & = e^{\ln x^{-5}} \\ & = x^{-5} \end{align}$
Thus, $\displaystyle zu = (1 - n) \int Qu \, dx + C$
$\displaystyle y^{-5}x^{-5} = -5 \int x^2 (x^{-5}) \, dx + C$
$\displaystyle x^{-5}y^{-5} = -5 \int x^{-3} \, dx + C$
$x^{-5}y^{-5} = -5 \left( \dfrac{x^{-2}}{-2} \right) + C$
$x^{-5}y^{-5} = \frac{5}{2} x^{-2} + C$
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Part (a) xdy + ydx = x^3 y^6 dx
$x \, dy + y \, dx = x^3 y^6 \, dx$
$dy + \dfrac{1}{x}y \, dx = x^2 y^6 \, dx$
$Q = x^2$
$n = 6$
$(1 - n) = -5$
$z = y^{1 - n} = y^{-5}$
Integrating Factor
$\begin{align}
\displaystyle u & = e^{(1 - n)\int P\,dx} \\
& = e^{-5\int \frac{1}{x}\,dx} \\
& = e^{-5 \ln x} \\
& = e^{\ln x^{-5}} \\
& = x^{-5}
\end{align}$
Thus,
$\displaystyle zu = (1 - n) \int Qu \, dx + C$
$\displaystyle y^{-5}x^{-5} = -5 \int x^2 (x^{-5}) \, dx + C$
$\displaystyle x^{-5}y^{-5} = -5 \int x^{-3} \, dx + C$
$x^{-5}y^{-5} = -5 \left( \dfrac{x^{-2}}{-2} \right) + C$
$x^{-5}y^{-5} = \frac{5}{2} x^{-2} + C$
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