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d-2-y-dx-2-tan-x-dy-dx-0-




Question Number 175406 by princeDera last updated on 29/Aug/22
(d^2 y/dx^2 ) −tan (x)(dy/dx) = 0
d2ydx2tan(x)dydx=0
Answered by Ar Brandon last updated on 29/Aug/22
y′′−tanx∙y′=0  ⇒((y′′)/(y′))=tanx  ⇒ln(y′)=−ln(cosx)+C  ⇒(dy/dx)=(1/(cosx))e^C =(k/(cosx))  ⇒y=∫ksecxdx=kln∣secx+tanx)+k_2
ytanxy=0yy=tanxln(y)=ln(cosx)+Cdydx=1cosxeC=kcosxy=ksecxdx=klnsecx+tanx)+k2

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