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The solution of differential equations sec2 y |
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Answer» iven, sec 2 x⋅tanydx+sec 2 y⋅tanxdy=0 On separating the varaibales, we get ⇒sec 2 x⋅tanydx=−sec 2 y⋅tanxdy ⇒ TANX sec 2 x DX=− TANY sec 2 y dy On integration both the SIDES, we get ∫ tanx sec 2 x dx=−∫ tany sec 2 y dy Let tanx=u⇒sec 2 x= dx DU ⇒dx= sec 2 x du and tany=v ⇒sec 2 y= dy dv ⇒dy= sec 2 y dv ∴∫ u sec 2 x ⋅ sec 2 x du =−∫ v sec 2 y ⋅ sec 2 y dv ⇒∫ u du =∫ v dv ⇒log∣u∣=−log∣v∣+log∣C∣ ⇒log∣tanx∣=−log∣tany∣+log∣C∣ ⇒log∣tanx⋅tany∣=log∣C∣ (∵logm+logn=logmn) (∵Step-by-step explanation: |
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