solHW 1_soln

solHW 1_soln - MATH150 Introduction to Ordinary...

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1 MATH150 Introduction to Ordinary Differential Equations Homework 1 Suggested Solution Prepared by CHAU Suk Ling, Pat 1. 2 2 ' x y yx y = + 2 2 2 20 2 2 2 0 2 2 2 0 2 2 2 2 (1 ) 2 1 2 1 1 ) 2 1 2l n ( 1 ) 2 n ( 2 2ln(1 ) 4 y x x dy x dx x ydy dx x x ydy dx x y dx x y x y x = + = + = + ⎡⎤ =+ ⎢⎥ + ⎣⎦ −= + + =− + + ∫∫ Note that we take the negative square root so that (0) 2 y = − . 2. 1 32 2 '( 1 ) y x 3 1 2 2 31 2 2 2 2 2 21 2 2 1 2 2 2 ) 1 ) 1 ) 11 1 1 ) 22 ) 1 2(1 ) dy x y dx x x dy dx y x x dy dx y x y x xC y = + = + = + −⋅ = ⋅ + + −=+ + Putting 0 x = and 1 y = , we have 1 2 2 2 1 2(1 0 ) 1 3 C C −= + +
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2 Thus 1 2 2 2 1 2 2 2 1 2 2 1 2(1 ) 3 1 32 ( 1 ) 1 ( 1 ) x y x y y x −=+ =− + = −+ Note that we take the positive square root so that (0) 1 y = . 3. 2 2 2 2 22 33 3 ' 34 3 (3 4) 3 (3 4) 3 4 x y y dy x dx y yd y x d x yx d x yy x C = = −= ∫∫ Putting 1 x = and 0 y = , we have 04 ( 0 )1 1 C C + =− Thus 41 x −=− . 4. 2 ' x e y y = + 00 0 0 2 0 2 2 (3 2 ) (2 ) (3 2 ) (2 ) 1 (3 2 ) (3 2 ) 2 2 1 (3 2 ) 2 1 4 19 (3 2 ) 2 1 44 384 1 3 2 x x x y x x y x x x dy e dx y ydy e dx y x e e e xe y = + += ⎡⎤ ++ = ⎣⎦ + +− = + + = Note that we take the positive square root so that (0) 0 y = . 5. With 2 2 t dx x te dt , an integrating factor is 2 2 dt t ee = . So we have 2 2 2 2 tt t t x et e e d t t d t t xe C =⋅ = =+
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3 Putting 1 t = and 0 x = , we have 1 2 C =− . Thus 2 2 2 2 1 2 1 2 t t t xe t x e = = 6. With 21 1 32 dx x t dt += , an integrating factor is 22 33 dt t ee = . So we have 222 333 1 (*) 2 tt t xe e te dt ⎛⎞ ⎜⎟ ⎝⎠ " . We first calculate 2 3 t te dt . Using integration by parts, we have 2 2 3 3 3 3 2 2 t t te dt tde te e dt te e == = ∫∫ Thus (*) becomes 2 3 2 3 1 2 31 3 3 2 2 393 284 21 3 (**) 84 t t t xe e te dt x t C xt C e tC e + =+− + =−+ " Note that for the graph to touch, but not to cross, the t -axis, we have 0 ,0 0 tt x dx dt = . Putting this to the given differential equation, we have 0 0 0( 0 ) 1 2 t t = This means the graph passes through (2, 0). Then (**) becomes 2 (2) 3 4 3 21 3 2 ) 9 8 Ce Ce =− + So the solution to the IVP is 42 21 3 9 84 8 t x te e =−− ⋅ and so 4 3 0 21 9 (0) 88 x xe . 7. With 13 s i n x −=+ i , an integrating factor is 1 dt t = . So we have (1 3sin ) 3 sin (*) t t xe t e dt e dt e tdt −− =+ = + " . We first calculate sin t et d t . Using integration by parts, we have
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4 sin (cos ) cos cos cos cos cos (sin ) cos sin sin cos sin sin 2s i n c o s s i n cos sin sin 2 tt t t t t et d te d t t d e d t e d t t d e d t d t d t −− =− + + = ∫∫
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solHW 1_soln - MATH150 Introduction to Ordinary...

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