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- 1. NOMBRE: ALEJANDRO CEVALLOS CARRERA: MECATRÓNICA ASIGNATURA: ECUACIONES DIFERENCIALES ORDIANARIA NRC: 4202 FECHA DE ENTREGA: 22-12-2020 DOCENTE: DR. JACQUELINE POZO TEMA: RESOLUCIÓN DE EJERCICIOS 1. 𝟐√𝒙𝒚′ − 𝒚 = −𝒔𝒆𝒏√𝒙 − 𝒄𝒐𝒔√𝒙 2√𝑥 𝑑𝑦 𝑑𝑥 − 𝑦 = −𝑠𝑒𝑛√𝑥 − 𝑐𝑜𝑠√𝑥 𝑑𝑦 𝑑𝑥 − 𝑦 2√𝑥 = − 𝑠𝑒𝑛√𝑥 2√𝑥 − 𝑐𝑜𝑠√𝑥 2√𝑥 𝑑𝑦 𝑑𝑥 − 𝑃(𝑥)𝑦 = 𝑄(𝑥) 𝑒∫𝑝(𝑥)𝑑𝑥 𝒆 − ∫ 1 2√𝑥 = 𝒆 − 1 2 ∫ 1 √𝑥 𝑑𝑥 = 𝑒 − 1 2 (2√𝑥) = 𝑒−√𝑥 𝑒−√𝑥 𝑦 = −∫ 𝑒−√𝑥 ( 𝑠𝑒𝑛√𝑥 2√𝑥 + 𝑐𝑜𝑠√𝑥 2√𝑥 )𝑑𝑥 𝑒−√𝑥 𝑦 = −(∫ 𝑒−√𝑥 𝑠𝑒𝑛√𝑥 2√𝑥 + ∫ 𝑒−√𝑥 𝑐𝑜𝑠√𝑥 2√𝑥 )𝑑𝑥 𝑢 = √𝑥; 𝑑𝑢 𝑑𝑥 = 1 2√𝑥 ; 𝑑𝑥 = 2√𝑥𝑑𝑢 −∫ 𝑒−√𝑥 𝑠𝑒𝑛√𝑥 2√𝑥 = ∫𝑒−𝑢 ( 𝑠𝑒𝑛𝑢 2𝑢 ) 2𝑢𝑑𝑢 = ∫ 𝑒−𝑢 𝑠𝑒𝑛 𝑢 𝑑𝑢 𝑡 = 𝑠𝑒𝑛𝑢; ∫𝑑𝑣 = ∫𝑒−𝑢 𝑑𝑢 𝑑𝑡 = 𝑐𝑜𝑠𝑢𝑑𝑢; 𝑣 = −𝑒−𝑢 ∫𝑒−𝑢 (𝑠𝑒𝑛𝑢)𝑑𝑢 = −𝑒−𝑢 𝑠𝑒𝑛𝑢 − ∫−𝑒−𝑢 (𝑐𝑜𝑠𝑢)𝑑𝑢
- 2. 𝑚 = 𝑐𝑜𝑠𝑢; ∫ 𝑑𝑛 = 𝑒𝑑𝑢 𝑑𝑚 = −𝑠𝑒𝑛𝑢𝑑𝑢; 𝑛 = −𝑒−𝑢 ∫ 𝑒−𝑢 𝑐𝑜𝑠𝑢𝑑𝑢 = − 𝑒−𝑢 𝑐𝑜𝑠𝑢 − ∫ −𝑒𝑢(𝑠𝑒𝑛𝑢)𝑑𝑢 ∫ 𝑒−𝑢 𝑐𝑜𝑠𝑢𝑑𝑢 = − 𝑒−𝑢 𝑐𝑜𝑠𝑢 − ∫ 𝑒𝑢(𝑠𝑒𝑛𝑢)𝑑𝑢 ∫ 𝑒−𝑢(𝑠𝑒𝑛 𝑢)𝑑𝑢 = −𝑒−𝑢 𝑠𝑒𝑛𝑢 − 𝑒−𝑢 cos𝑢 + ∫ ∫𝑒−𝑢 cos𝑢 2∫ 𝑒−𝑢 𝑠𝑒𝑛𝑢𝑑𝑢 = −𝑒−𝑢 𝑠𝑒𝑛 𝑢 − 𝑒−𝑢 cos𝑢 ∫ 𝑒−𝑢 𝑠𝑒𝑛 𝑢 𝑑𝑢 = − 𝑒−𝑢 𝑠𝑒𝑛 𝑢 + 𝑒−𝑢 cos𝑢 2 + 𝑐 − ∫ 𝑒−√𝑥 𝑐𝑜𝑠√𝑥 2√𝑥 𝑑𝑥 𝑢 = √𝑥 ;𝑑𝑢 = 1 2√𝑥 𝑑𝑥; 𝑑𝑥 = 2√𝑥𝑑𝑢 ∫ 𝑒−𝑢 cos𝑢 2𝑢 (2𝑢𝑑𝑢) = ∫ 𝑒−𝑢 cos𝑢 𝑑𝑢 ∫ 𝑒−𝑢 cos𝑢𝑑𝑢 = −𝑒𝑢 cos𝑢 − (−𝑒−𝑢 + ∫ 𝑒−𝑢 cos𝑢𝑑𝑢) ∫𝑒−𝑢 cos𝑢 𝑑𝑢 = −𝑒−𝑢 cos𝑢 − (−𝑒−𝑢 𝑠𝑒𝑛 𝑢 + ∫𝑒−𝑢 cos𝑢 𝑑𝑢) 2 ∫𝑒−𝑢 𝑐𝑜𝑠𝑢 𝑑𝑢 = −𝑒𝑢 cos𝑢 + 𝑒−𝑢 𝑠𝑒𝑛 𝑢 𝑒−√𝑥 𝑦 = − (− 𝑒−𝑢 + 𝑒−𝑢 cos𝑢 2 + 𝑒−𝑢 𝑠𝑒𝑛 𝑢 − 𝑒−𝑢 cos𝑢 2 ) + 𝑐 𝑒−√𝑥 𝑦 = 𝑒−𝑢 + 𝑒−𝑢 cos𝑢 2 − 𝑒−𝑢 𝑠𝑒𝑛 𝑢 − 𝑒−𝑢 cos𝑢 2 + 𝑐 2𝑒−√𝑥 𝑦 + 2𝑐 = 𝑒−𝑢 𝑠𝑒𝑛 𝑢 + 𝑒−𝑢 cos𝑢 − 𝑒−𝑢 𝑠𝑒𝑛 𝑢 + 𝑒−𝑢 cos𝑢 + 𝑐 2(𝑒−√𝑥 + 𝑐) = 2𝑒−𝑢 cos𝑢 𝑒−√𝑥 𝑦 + 𝑐 = 𝑒−𝑢 cos 𝑢 𝑦 = 𝑒−√𝑥 𝑐𝑜𝑠√𝑥 𝑒−√𝑥 + 𝑐 𝑒−√𝑥
- 3. 𝒚 = 𝒄𝒐𝒔√𝒙 + 𝒄 ∗ 𝒆√𝒙 2. (𝒙𝒚𝟐 + 𝒚)𝒅𝒙 − 𝒙𝒅𝒚 = 𝟎) (𝑥𝑦2 + 𝑦)𝑑𝑥 = 𝑥𝑑𝑦 𝑑𝑦 𝑑𝑥 = 𝑥𝑦2 + 𝑦 𝑥 = 𝑦2 + 𝑦 𝑥 𝑦′ − 𝑦 𝑥 = 𝑦2 = 𝑦−2 𝑦′ − 1 𝑥𝑦 = 1 𝑧 = 𝑦−1 𝑑𝑧 𝑑𝑥 = −𝑦−2 ( 𝑑𝑦 𝑑𝑥 ) −𝑦2 ( 𝑑𝑧 𝑑𝑥 ) = 𝑑𝑦 𝑑𝑥 𝑦−2 (−𝑦2 𝑑𝑧 𝑑𝑥 ) − 1 𝑥𝑦 = 1 − 𝑑𝑧 𝑑𝑥 − 𝑧 𝑥 = 1 𝑑𝑧 𝑑𝑥 + 𝑧 𝑥 = −1 𝑒√𝑥 = 𝑒ln(𝑥) ;𝑒𝑑𝑥 = 𝑥 𝑧𝑥 = − ∫𝑥𝑑𝑥 𝑧𝑥 = − 𝑥2 2 + 𝑐 𝑧 = − 𝑥2 2 + 𝑐 𝑥 𝑧 = −𝑥2 + 𝑐 2𝑥 1 𝑦 = −𝑥2 + 𝑐 2𝑥 𝒚 = 𝟐𝒙 −𝒙𝟐 + 𝒄
- 4. 3. (𝒙 − 𝟐𝒙𝒚 − 𝒚𝟐)𝒚′ + 𝒚𝟐 = 𝟎 (𝑥 − 2𝑥𝑦 − 𝑦2)𝑑𝑦 + 𝑦2 𝑑𝑥 = 0 (𝑥 − 𝑥𝑦 − 𝑥𝑦 − 𝑦2)𝑑𝑦 + 𝑦2 𝑑𝑥 = 0 [𝑥(1− 𝑦) − 𝑥𝑦(1 − 𝑦)]𝑑𝑦 + 𝑦2 𝑑𝑥 = 0 𝑥(1 − 𝑦)(1 − 𝑦)𝑑𝑦 + 𝑦2 𝑑𝑥 = 0 𝑥(1 − 𝑦)2 𝑑𝑦 = −𝑦2 𝑑𝑥 (1 − 𝑦)2 𝑦2 𝑑𝑦 = − 1 𝑥 𝑑𝑥 1 − 2𝑦 + 𝑦2 𝑦 𝑑𝑦 = − 1 𝑥 𝑑𝑥 ∫( 1 𝑦2 − 2 𝑦 + 1)𝑑𝑦 = ∫ − 1 𝑥 𝑑𝑥 − 1 𝑦 − 2ln|𝑦| − 𝑦 + 𝑐 = − ln|𝑥| ln|𝑥| = 1 𝑦 + 2 ln|𝑦| − 𝑦 + 𝑐 𝑒ln|𝑥| 𝑒𝑐 = 𝑒𝑦−1 + 𝑒ln|𝑦2|−𝑦 𝑥𝑐 = 𝑒𝑦−1 𝑦2 𝒚𝟐 = 𝒙𝒄 𝒆𝒚−𝟏 4. 𝒙𝟐 𝒚𝒏 𝒚′ = 𝟐𝒙𝒚′ − 𝒚, 𝒏 ≠ −𝟐 𝑥2 𝑦𝑛 𝑦′ − 2𝑥𝑦′ + 𝑦 = 0 𝑦′(𝑥2 𝑦𝑛 − 2𝑥) + 𝑦 = 0 𝑑𝑥 𝑑𝑦 + 𝑦 𝑥2𝑦𝑛 − 2𝑥 = 0 𝑑𝑥 𝑑𝑦 + (𝑥2 𝑦𝑛 − 2𝑥) 𝑦 = 0 𝑑𝑥 𝑑𝑦 + 𝑥2 𝑦𝑛−1 − 2𝑥 𝑦 = 0 1 𝑥2 ( 𝑑𝑥 𝑑𝑦 ) + 𝑦𝑛−1 − 2 𝑥𝑦 = 0
- 5. 𝑧 = 1 𝑥 ; 𝑑𝑧 𝑑𝑦 = − 1 𝑥2 𝑑𝑥 𝑑𝑦 ; −𝑥2 𝑑𝑧 𝑑𝑦 = 𝑑𝑥 𝑑𝑦 − 𝑑𝑧 𝑑𝑦 − 2 𝑦 𝑧 = 0 𝑑𝑧 𝑑𝑦 = − 2𝑧 𝑦 ∫ 𝑑𝑧 𝑧 = −∫ 2 𝑦 𝑑𝑦 ln|𝑧| = −2ln|𝑦| + 𝑐 𝑒ln|𝑧| = 𝑒ln|𝑦−2| + 𝑒𝑐 𝒛 = 𝒚−𝟐 + 𝒆𝒄 5. 𝒚𝒄𝒐𝒔𝒙𝒅𝒙 + (𝟐𝒚𝒔𝒆𝒏𝒙)𝒅𝒚 = 𝟎 𝑢 = 𝑠𝑒𝑛 𝑥 ;𝑦𝑑𝑢 = cos 𝑥𝑑𝑥 𝑦𝑑𝑢 + (2𝑦 − 𝑢)𝑑𝑦 = 0 𝑑𝑦 𝑑𝑢 = 𝑦 𝑢 − 2𝑦 𝑑𝑦 𝑑𝑢 = 𝑦 𝑢 𝑦 𝑢 − 2𝑦 𝑢 𝑡 = 𝑦 𝑢 ; 𝑢𝑡 = 𝑦 ;𝑡 + 𝑢𝑑𝑡 𝑑𝑢 = 𝑑𝑦 𝑑𝑢 𝑑𝑦 𝑑𝑢 = 𝑦 𝑢 1 − 2𝑦 𝑢 𝑡 + 𝑢𝑑𝑡 𝑑𝑢 = 𝑡 1 − 2𝑡 𝑢 𝑑𝑡 𝑑𝑢 = 𝑡 1 − 2𝑡 − 𝑡 𝑢 𝑑𝑡 𝑑𝑢 = 𝑡 − 𝑡 + 2𝑡2 1 − 2𝑡 𝑢 𝑑𝑡 𝑑𝑢 = 2𝑡2 1 − 2𝑡
- 6. 1 − 2𝑡 𝑡2 𝑑𝑡 = 𝑑𝑢 𝑢 ∫ ( 1 𝑡2 − 2 𝑡 ) 𝑑𝑡 = ∫ 𝑑𝑢 𝑢 − 1 𝑡 − 2 ln|𝑡| + 𝑐 = ln|𝑢| 𝑒−𝑡−1 − 𝑡2 + 𝑒𝑐 = 𝑢 𝑒 − 𝑦 𝑢 − ( 𝑦 𝑢 ) 2 + 𝑒𝑐 = 𝑠𝑒𝑛 𝑥 𝑒 − 𝑠𝑒𝑛𝑥 𝑦 − 𝑦2 𝑠𝑒𝑛2 𝑥 + 𝑒𝑐 = 𝑠𝑒𝑛𝑥 𝒆 − 𝒔𝒆𝒏𝒙 𝒚 + 𝒆𝒄 = 𝒔𝒆𝒏𝒙 + 𝒚𝟐 𝒔𝒆𝒏𝟐𝒙