Question

b) 2 - th(y)+ Solve the following differential equations: dy 5 + 2x a) 1 - v3y dy c) ; y(2) = 0 dx y 21 dy 1 + y d) My(0) = -1 2 dt 1 + 2 y dx 1 + x dy dy y e) X + 2 y = x'; x > = x'; x > 0; y(1) = 0 f) X = x; x > 0; y(1) = 0 dx dx x + 1

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Answer 1

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dy dx 5+ (2x but to all using variable seperable method - ☺ (1-3y) dy = (5+ să de f ( 1 - 33 y 3 ) dy =(( 5 + 22 22 ) da fidy - gh fys dy = (s. dx + √2 fezca y - ok. yyzt! (x+1) 5x + 5z. x Yet | + C. (Y2+1) y - 33 а 473 52t 52 x 3/2 tc (32) y - 33.3 4 4/3 5x+ 52.2 x 3 te y- ab a gy's - sz. f.22.2% to y - 34.834 5x + 2 x √ 2x + c = 5x + 3 x x + 2 x J 22 te (Arg)

PAGE: (6) h tin (yet) + t² och using Properties In (m) = nlnm nou ih h = t (24) Iny+ t² عم عم عمل 2t² lng + t t² ( 2 y + 1) Y' y y'= dyat dy : I ² (2 ling+1) d* y dy tedt y (2 my+1) (variable seperable) f y (2 ny+1) - (+B) reystitute alny+ 1 = u 2. z dy = du = Idy. 3 batan tara dy 등 ц 3 I Infly = +3 3 2 + c 1 lb / 2 lng+11 3 Ans

(c) dy = dt ; y (2) = 0 Itzy Using variable seperable method- (Itzy) dy at at 1 (1+2y) dy et dt + c 2 Y+ 2.4² + Y+ y² 2 t² + c 0+ y (2)zº, jieif t=2; 920 = (2)²+ c 7c = -4 y + y² = t - 4. y (4+1) = +²4 (Any), dy T + y 2 yolal Ita² using variable seperable method dy वx +y² Itx² dx eb -T dx le x² ity2

tanly tantate y col=-1, ie, if azo, Y = -1 = ) tan (-1) = - tant (o)tc = - = -(0) + c Ca - Aly tany tanx - 7/4 Any