Question Solved1 Answer Solve the initial value problem y' = 4y2 + xy2, y(0) = 1 and determine where the solution attains its minimum value. Enclose numerators and denominators in parentheses. For example, (a - b)/(1+n). y(x) = Qe Minimum value: x = Number

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Transcribed Image Text: Solve the initial value problem y' = 4y2 + xy2, y(0) = 1 and determine where the solution attains its minimum value. Enclose numerators and denominators in parentheses. For example, (a - b)/(1+n). y(x) = Qe Minimum value: x = Number
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Transcribed Image Text: Solve the initial value problem y' = 4y2 + xy2, y(0) = 1 and determine where the solution attains its minimum value. Enclose numerators and denominators in parentheses. For example, (a - b)/(1+n). y(x) = Qe Minimum value: x = Number
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Solve the equation and than minimise the resulting equation. Answer fo y^(')=4y^(2)+xy^(2)quad y(0)=1.{:[dy//dx=4y^(2)+xy^(2)=dy//dx=y^(2)(4+x).],[dy//y^(2)=(4+x)dx.]:}on iitegrating twis we get1-1//y=4x+(x^(2))/(2)+c∣Now we have y(0)=$. so wing it we can find Cputring x=0 we cam have y=2.so we can g ... See the full answer