Browse Questions

# Solution of differential equation $\large\frac{dy}{dx} +\frac{y}{x}$$=x^3 y^5 is which of the following? (a)\;(x-y)^4=-4 \log x +c \\ (b)\;(xy)^{-5} =\frac{5}{6}x^{-2} +c \\ (c)\;(xy)^{-5} =\log x +c \\ (d)\;(xy)^{-4}=-4 \log x +c Can you answer this question? ## 1 Answer 0 votes \large\frac{1}{y^5}\frac{dy}{dx}+\frac{1}{xy^4}$$=x^3$
$y^{-4}=v$
$-4y^{-5} \large\frac{dy}{dx} =\large\frac{dv}{dx}$
equation becomes $\large\frac{-1}{4} \times \frac{dv}{dx}+\frac{v}{x}$$=x^3 \large\frac{dv}{dx} -\frac{4v}{x}$$=-4x^3$-----(i)
This linear differential equation $I.F= e^{\int \large\frac{4}{x} dx}$
$\qquad= e^{-4 \log x}=\large\frac {1}{x^4}$
Multiplying both sides by (i) by I.F we get
$v. \large\frac{1}{x^4}$$=\int -4x^3. \large\frac{1}{x^4}dx \large\frac {v}{x^4}$$=-4 \log x +c$
$x^4y^{-4}=-4 \log x +c$
Hence d is the correct answer.