Author Topic: TUT 0702 QUIZ1  (Read 612 times)

Kunpeng Liu

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TUT 0702 QUIZ1
« on: September 27, 2019, 02:29:41 PM »
$$
Find\, \, \, \,the\, \, \, solution\, \, of\, \,  the\, \, \,  given\, \,  initial\, \,  value\, \,  problem\, \,  in\, \,  explicit\, \,  form:{y}'=2x/(1+2y), y(2)=0\\\\\frac{dy}{dx}=\frac{2x}{1+2y}\\\\(1+2y)dy=2xdx\\\\\int (1+2y)dy=\int 2xdx\\\\(y+1/2)^{2}=x^{2}+c\\\\y+\frac{1}{2}=(x+c)^{1/2}\\\\y=(x+c)^{1/2}-\frac{1}{2}\\\\\because y(2)=0 \rightarrow 0=(4+c)^{1/2}-\frac{1}{2}\\\\therefore\, \, C=\frac{-15}{16}\\\\y=(x^{2}-\frac{15}{16})^{1/2}-\frac{1}{2}
$$
« Last Edit: September 27, 2019, 10:05:41 PM by Kunpeng Liu »

Kunpeng Liu

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Re: TUT 0702 QUIZ1
« Reply #1 on: September 27, 2019, 02:32:21 PM »
$$
Find\, \, \, \,the\, \, \, solution\, \, of\, \,  the\, \, \,  given\, \,  initial\, \,  value\, \,  problem\, \,  in\, \,  explicit\, \,  form:{y}'=2x/(1+2y), y(2)=0\\\\\frac{dy}{dx}=\frac{2x}{1+2y}\\\\(1+2y)dy=2xdx\\\\\int (1+2y)dy=\int 2xdx\\\\(y+1/2)^{2}=x^{2}+c\\\\y+\frac{1}{2}=(x+c)^{1/2}\\\\y=(x+c)^{1/2}-\frac{1}{2}\\\\\because y(2)=0 \rightarrow 0=(4+c)^{1/2}-\frac{1}{2}\\\\therefore\, \, C=\frac{-15}{16}\\\\y=(x^{2}-\frac{15}{16})^{1/2}-\frac{1}{2}
$$
« Last Edit: September 27, 2019, 10:06:11 PM by Kunpeng Liu »