Shazia Abbas

(𝒳 2 + 𝒴 2 )𝒹𝒳 + (𝒳 2 − 𝒳𝒴 )𝒹𝒴 = 0 → (𝑖 ) ; )2 C1(𝑥 + 𝑦 =X℮ Derivative 𝑦 𝑥 w.r.t ‘x’ 2C1(𝑥 + 𝑦) (1 + 2( 𝑦 𝑋℮𝑥 𝑥+𝑦) 𝑑𝑦 𝑑𝑥 𝑦 𝑥 𝑑𝑦−𝑦 𝑑𝑥 𝑥 𝑥2 𝑦 𝑥 )=℮ +x℮ ( ( ) 2 𝑥 + 𝑦 (1 + 𝑑𝑦 𝑑𝑥 )=℮ 𝑦 𝑥 ) 𝑦 𝑥 𝑑𝑦−𝑦 𝑑𝑥 𝑥 𝑥2 +x℮ ( 𝑌 " = 𝑌 ; 𝑌 = cos ℎ𝑥 + sin ℎ𝑥 Y"=Y→ (𝑖 ) ; 𝑌 = 𝑐𝑜𝑠 ℎ𝑥 + 𝑠𝑖𝑛 ℎ𝑥 Derivative w.r.t ‘x’ 𝑌 = 𝑐𝑜𝑠 ℎ𝑥 + 𝑠𝑖𝑛 ℎ𝑥 𝑌 ′= 𝑠𝑖𝑛 ℎ𝑥 + 𝑠𝑖𝑛 ℎ𝑥 𝑌 " = 𝑐𝑜𝑠 ℎ𝑥 + 𝑠𝑖𝑛 ℎ𝑥 Values of 𝑌 " 𝑎𝑛𝑑 𝑌 put in (𝑖 ), we have 𝑐𝑜𝑠 ℎ𝑥 + 𝑠𝑖𝑛 ℎ𝑥 = 𝑐𝑜𝑠 ℎ𝑥 + 𝑠𝑖𝑛 ℎ𝑥 ) 1=1→ (𝑖𝑖 ) Equation (𝑖𝑖 ) 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (𝑖 ) 𝑌'' + 25𝑌 = 0 ; 𝑌 = 𝐶1 cos 5𝑥 𝑌'' + 25𝑌 = 0 → (𝑖 ) ; 𝑌 = 𝐶1 𝑐𝑜𝑠 5𝑥 Derivative w.r.t ‘x’ 𝑌 = 𝐶1 𝑐𝑜𝑠 5𝑥 Question: 1 𝑑𝑦 𝑥 2 𝑦 2 = 𝑑𝑥 1 + 𝑥 Solution: 𝑑𝑦 𝑥 2 𝑦 2 = 𝑑𝑥 1 + 𝑥 (1 + 𝑥 ) 𝑑𝑥 =𝑥 2 𝑦 2 𝑑𝑦 1+𝑥 ( 2 ) 𝑑𝑥 = 𝑦 2 𝑑𝑦 𝑥 1 1 ∫ ( 2 + ) 𝑑𝑥 = ∫ 𝑦 2 𝑑𝑦 𝑥 𝑥 ∫𝑥 −2 𝑑𝑥 + ∫ 1 𝑥 1 𝑑𝑥 = 𝑦 3 3 −1 1 3 + log|𝑥 | + 𝑐 = 𝑦 𝑥 3 Or -3+3x𝑙𝑜𝑔|𝑥| + 3𝑐𝑥 = 𝑥 𝑦 3 -3+3x𝑙𝑜𝑔|𝑥| + 𝑐1 𝑥 = 𝑥 𝑦 3 Question: 2 2y(𝑥 + 1)𝑑𝑦 = 𝑥𝑑𝑥 Solution: 𝑥 2y𝑑𝑦=𝑥+1 𝑑𝑥 𝑥+1−1 ∫ 2𝑦𝑑𝑦 = ∫ 𝑑𝑥 (1 + 𝑥 ) 1 ∫ 2𝑦𝑑𝑦 = ∫ (1 − ) 𝑑𝑥 1+𝑥 1 2 ∫ 𝑦𝑑𝑦 = ∫ 1 𝑑𝑥 − ∫ 𝑑𝑥 1+𝑥 𝑦 2 = 𝑥-log |1 + 𝑥|+c 𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛: 1 𝑡𝑎𝑛 𝑥 , 𝑐𝑜𝑡 𝑥 , 𝜋 (𝑜, ) 2 Solution: 𝑓1(𝑥) =tan 𝑥 , 𝑓2(𝑥) =𝑐𝑜𝑡 𝑥 We know that w(𝑓1(𝑥) , 𝑓2(𝑥) )=0 w(𝑡𝑎𝑛 𝑥 , 𝑐𝑜𝑡 𝑥)=| 𝑡𝑎𝑛 𝑥 𝑠𝑒𝑐 2 𝑥 𝑐𝑜𝑡 𝑥 | −𝑐𝑜𝑠𝑒𝑐 2 𝑥 = −𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑡𝑎𝑛 𝑥 − 𝑠𝑒𝑐 2 𝑥 𝑐𝑜𝑡 𝑥 1 sin 𝑥 1 cos 𝑥 =− 𝑠𝑖𝑛2 𝑥 cos 𝑥 − 𝑐𝑜𝑠2𝑥 sin 𝑥 =− cosec 𝑥 sec 𝑥 − 𝑠𝑒𝑐 𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 =−2 𝑠𝑒𝑐 𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 ≠ 0 for 0< 𝑥 < 𝜋 2 Question: 2 ℮𝑥 , ℮−𝑥 , ℮4𝑥 ; (−∞, ∞) Solution: 𝑓1(𝑥) = ℮𝑥 , 𝑓2(𝑥) = ℮−𝑥 , 𝑓3(𝑥) = ℮4𝑥 We know that w(𝑓1(𝑥) , 𝑓2(𝑥), 𝑓3(𝑥) )=0 ℮𝑥 w(℮𝑥 , ℮−𝑥 , ℮4𝑥 )=|℮𝑥 ℮𝑥 ℮−𝑥 −℮−𝑥 ℮−𝑥 ℮4𝑥 4℮4𝑥 | 8℮4𝑥 =[ ℮𝑥 (−16℮3𝑥 − 4℮3𝑥 ) − ℮−𝑥 (16℮5𝑥 − 4℮5𝑥 ) + ℮4𝑥 (1 + 1)] =[℮𝑥 (−20℮3𝑥 ) − ℮−𝑥 (12℮5𝑥 ) + 2℮4𝑥 ] =−20℮4𝑥 − 12℮4𝑥 + 2℮4𝑥 =−30℮4𝑥 ≠ 0 𝑓𝑜𝑟 − ∞ < 𝑥 < ∞