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Question Number 59926 by rahul 19 last updated on 16/May/19

Commented by rahul 19 last updated on 16/May/19

$$5,6.$$

Answered by tanmay last updated on 16/May/19

$$cosxdy=ysinxdx-y^{2}dx$$$$-ysinxdx+cosxdy=-y^{2}dx$$$$\frac{-ysinxdx+cosxdy}{1}=-y^{2}dx$$$$d\left(cosx\times y\right)=-y^{2}dx$$$$\frac{d\left(ycosx\right)}{y^2{cos}^{2}x}=-\frac{dx}{{cos}^{2}x}$$$$\frac{-1}{\left(ycosx\right)}=-tanx-c$$$$secx=y(tanx+c)$$$$$$$$$$

Answered by tanmay last updated on 16/May/19

$$\frac{dy}{dx}=\frac{{xy}^5+2y}{x}$$$$\frac{dy}{dx}=y^5+\frac{2y}{x}$$$$\frac{dy}{dx}-\frac{2y}{x}=y^5$$$$y^{-5}\frac{dy}{dx}-\frac{2y^{-4}}{x}=1$$$$t=y^{-5+1}$$$$\frac{dt}{dx}=-4\times y^{-5}\times \frac{dy}{dx}$$$$\frac{-1}{4}\times \frac{dt}{dx}-\frac{2t}{x}=1$$$$\frac{dt}{dx}+\frac{8t}{x}=-4$$$$Intregatingfactor{e}^{\int \frac{8}{x}dx}=e^{8lnx}=x^8$$$$x^8\frac{dt}{dx}+8x^{7}t=-4x^8$$$$\frac{d}{dx}\left(x^8\times t\right)=-4x^8$$$$d\left(x^{8}t\right)=-4x^{8}dx$$$$\int d\left(x^{8}t\right)=-4\int x^{8}dx$$$$x^{8}t=\frac{-4}{9}\times x^9+c$$$$x^8\times y^{-4}=\frac{-4}{9}\times x^9+c$$$$9x^8=-4x^9{y}^4+y^4\times 9c$$$$9x^8+4x^9{y}^4=y^4\times 9c$$

Commented by rahul 19 last updated on 16/May/19

$$thankyousir!$$

Commented by tanmay last updated on 16/May/19

$$mostwelcome...nowithinkyouarecounting$$$$daystowrestlewiththemath+physicsin$$$$comingexam...$$

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