# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: Kunpeng Liu on September 27, 2019, 02:29:41 PM

Title: TUT 0702 QUIZ1
Post by: Kunpeng Liu on September 27, 2019, 02:29:41 PM
$$Find\, \, \, \,the\, \, \, solution\, \, of\, \, the\, \, \, given\, \, initial\, \, value\, \, problem\, \, in\, \, explicit\, \, form:{y}'=2x/(1+2y), y(2)=0\\\\\frac{dy}{dx}=\frac{2x}{1+2y}\\\\(1+2y)dy=2xdx\\\\\int (1+2y)dy=\int 2xdx\\\\(y+1/2)^{2}=x^{2}+c\\\\y+\frac{1}{2}=(x+c)^{1/2}\\\\y=(x+c)^{1/2}-\frac{1}{2}\\\\\because y(2)=0 \rightarrow 0=(4+c)^{1/2}-\frac{1}{2}\\\\therefore\, \, C=\frac{-15}{16}\\\\y=(x^{2}-\frac{15}{16})^{1/2}-\frac{1}{2}$$
Title: Re: TUT 0702 QUIZ1
Post by: Kunpeng Liu on September 27, 2019, 02:32:21 PM
$$Find\, \, \, \,the\, \, \, solution\, \, of\, \, the\, \, \, given\, \, initial\, \, value\, \, problem\, \, in\, \, explicit\, \, form:{y}'=2x/(1+2y), y(2)=0\\\\\frac{dy}{dx}=\frac{2x}{1+2y}\\\\(1+2y)dy=2xdx\\\\\int (1+2y)dy=\int 2xdx\\\\(y+1/2)^{2}=x^{2}+c\\\\y+\frac{1}{2}=(x+c)^{1/2}\\\\y=(x+c)^{1/2}-\frac{1}{2}\\\\\because y(2)=0 \rightarrow 0=(4+c)^{1/2}-\frac{1}{2}\\\\therefore\, \, C=\frac{-15}{16}\\\\y=(x^{2}-\frac{15}{16})^{1/2}-\frac{1}{2}$$