3x2y+x3dydx=3cos x⇒x3y=

Given dydx=3 cosx−3x2yx3⇒ dydx=3cosxx3−3yx⇒ dydx+3x.y= 3cosxx3I.F= e∫3x dx=e3 logx=e log x3=x3Solution is y.(I.F)= ∫3 cosxx3 (I.F)⇒ y.x3= ∫3cosxx3.x3 dx⇒ y.x3=3 sinx+c

3x2y+x3dydx=3cos x⇒x3y=

Given dydx=3 cosx−3x2yx3⇒ dydx=3cosxx3−3yx⇒ dydx+3x.y= 3cosxx3I.F= e∫3x dx=e3 logx=e log x3=x3Solution is y.(I.F)= ∫3 cosxx3 (I.F)⇒ y.x3= ∫3cosxx3.x3 dx⇒ y.x3=3 sinx+c

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$3 x^{2} y + x^{3} \frac{dy}{dx} = 3 \cos x \Rightarrow x^{3} y =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

sin x

3cos x + c

2sin x + c

3sin x + c

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} G i v e n \frac{d y}{d x} = \frac{3 \cos x - 3 x^{2} y}{x^{3}} \\ \Rightarrow \frac{d y}{d x} = \frac{3 \cos x}{x^{3}} - \frac{3 y}{x} \\ \Rightarrow \frac{d y}{d x} + \frac{3}{x} . y = \frac{3 \cos x}{x^{3}} \\ I . F = e^{\int \frac{3}{x} d x} = e^{3 \log x} = e^{\log x^{3}} = x^{3} \\ S o l u t i o n i s y . (I . F) = \int \frac{3 \cos x}{x^{3}} (I . F) \\ \Rightarrow y . x^{3} = \int \frac{3 \cos x}{x^{3}} . x^{3} d x \\ \Rightarrow y . x^{3} = 3 \sin x + c \end{array}$