Question

(15 points) The equation 5xy + 2x2 + 2y2 5y (*) .22 can be written in the form y' = f(y/2), i.e., it is homogeneous, so we can use the substitution u=y/2 to obtain a separable equation with dependent variable u = u(x). Introducing this substitution and using the fact that y' = ru' + u we can write (*) as y' = xu'+u = f(u) where f(u) = Separating variables we can write the equation in the form du g(u) du where g(u) = An implicit general solution with dependent variable u can be written in the form In(2)- =C. Transforming u = y/x back into the variables x and y and using the initial condition y(1) = 1 we find C= Finally solve for y to obtain the explicit solution of the initial value problem Y

Answer 1

Ques sy' = 5xy + 2x² + 2y 2 22 using U=.y|x y=uu → y'= autu Putting t_glo in o 5y's 5y we have + 59' = 5 uta+què y = u + 2 + 24² . file) ut a 5 +

.. y'= xuru= ut u + 2 + 24² xu = stau ² 5 x du 2 (1+2) da 5 1 5 du 1+42 a da 9 5 11 du da & Lite) &(142) grdegrating both sides of ① tantu+E=lnx Ina. tanu uc putting u= yle, we have Inn 3 tan 16 - C 2

As y (1) =1 In 1 - 5 tan tant (t) = - 5 tan (1) = 6 stan (1) = 52 I - C 2 a C = – 5 tanel) ou -51 버 2 1 o ina- 5 tart ( 4 ) = -g tanter) tar ( lax + 5 tant(1) 2 to tant (1) 요 x tan & lnx tta dax+ tanti) Ha510) an (& lnx + 1) ac 11 x tan