Question

If $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x\Rightarrow x^3{y}^{-3}=$

• A. $\sin x$
• B. $2\sin x+c$
• C. $3\sin x+c$
• D. $3\cos x+c$

Answer 1

Answer: (C)

$3\sin x+c$

$x^{2}y-x^3\frac{dy}{dx}=y^4\cos xdividingby-x^{3}onbothsides\frac{dy}{dx}-\frac{y}{x}=\frac{-y^4}{x^3}\cos x\Rightarrow \frac{1}{y^4}\frac{dy}{dx}-\frac{1}{y^{3}x}=-\frac{1}{x^3}\cos xputting\frac{1}{y^3}=t\frac{-3}{y^4}\frac{dy}{dx}=\frac{dt}{dx}\Rightarrow \frac{-1}{3}\frac{dt}{dx}-\frac{t}{x}=-\frac{-1}{x^3}\cos x\Rightarrow \frac{dt}{dx}+\frac{3t}{x}=\frac{3}{x^3}\cos xitisafirstdegreelinearD.Eherep=\frac{3}{x}Q=\frac{3}{x^3}\cos xI.F=e^{\int pdx}=e^{\int \frac{3}{x}dx}=e^{3\log x}=x^{3}solutionofD.Eist\times I.F=\int Q.I.Fdx+Ct\times x^3=\int \frac{3}{x^3}\cos x\times x^{3}dx+C\Rightarrow \frac{x^3}{y^3}=\int 3\cos xdx+C\Rightarrow x^3{y}^{-3}=3\sin x+C$