### Author Topic: TUT 0702 QUIZ1  (Read 788 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

• Jr. Member
• Posts: 9
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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 »