# Diff Eqns - = 3 1 y = 16 y 5 e 1 dt dy = 5 4 y = 17 y 4 e 1 dt dy = 4 7 y-= 1 x 3 e 2 y-= 2 t 2 e 10 y = 3 2 x 2 e 5 y = 4 t e 4 y-= 5 x e 14 y-= 6

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Name ____________________________ Date ____________________________ Period ____________________________ Calculus B Solve for y by the method of separation of variables. 1. y 3 dx dy - = ; y(0) = 2 2. y 2 dt dy = ; y(0) = 10 3. 0 xy 4 dx dy = - ; y(0) = 5 4. 0 y dt dy = + ; y(0) = 4 5. 0 y dx dy = + ; y(0) = 14 6. ky dt dy = , k is a nonzero constant; y(0) = 1 7. ky dx dy = , k is a nonzero constant; y(0) = 7 8. 32 y 2 dt dy - - = ; y(0) = -50. Hint: Factor the right hand side first, find y, then find C. 9. 15 y 3 dt dy - - = ; y(0) = -20. Hint: Factor the right hand side first, find y, then find C. 10. y e x dx dy = ; y(0) = 1 11. y e x 2 dx dy = ; y(0) = 2 12. y 3 e t dt dy - = ; y(0) = -2 13. y e 1 dt dy - = ; y(0) = 4 14. y 2 2 e x 3 dx dy = ; 2 1 ) 0 ( y = 15. y 3 e x 2 dx dy

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Unformatted text preview: = ; 3 1 ) ( y = 16. y 5 e 1 dt dy = ; 5 4 ) ( y = 17. y 4 e 1 dt dy = ; 4 7 ) ( y-= 1. x 3 e 2 y-= 2. t 2 e 10 y = 3. 2 x 2 e 5 y = 4. t e 4 y-= 5. x e 14 y-= 6. kt e y = 7. kx e 7 y = 8. 16 e 34 y t 2--=-9. 5 e 15 y t 3--=-10. + = e x 2 1 ln y 2 11. ( 29 2 2 e x ln y + = 12. +--= 2 4 e t 4 1 ln y 13. ( 29 4 e t ln y-+--= 14. ( 29 e x 2 ln 2 1 y 3 + = 15. ( 29 e x 3 ln 3 1 y 2 + = 16. ( 29 4 e t 5 ln 5 1 y + = 17. ( 29 7 e t 4 ln 4 1 y-+ =...
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## This note was uploaded on 02/21/2010 for the course MATH 1A/Ma taught by Professor (all) during the Spring '10 term at University of California, Berkeley.

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Diff Eqns - = 3 1 y = 16 y 5 e 1 dt dy = 5 4 y = 17 y 4 e 1 dt dy = 4 7 y-= 1 x 3 e 2 y-= 2 t 2 e 10 y = 3 2 x 2 e 5 y = 4 t e 4 y-= 5 x e 14 y-= 6

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