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# 681014.docx - Running head DEFERENTIAL SOLUTIONS...

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Running head: DEFERENTIAL SOLUTIONS Deferential Solutions Name of the University: Name of the Student: Author Note

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DEFERENTIAL SOLUTIONS Question 1: Find the explicit general solution of xdx - (y^2)dy=0. Xdx - (y^2)dy = 0 Or, xdx = (y^2)dy integrating both sides we get, (x^2)/2 + C1 = (y^3)/3 + C2 Or, 3(x^2) + 3C1 - 2(y^3) - 2C2 = 0 Let, 3C1 - 2C2 = C’ Therefore, 3(x^2) - 2(y^3) + C’ = 0 Hence, this is the general solution of the equation. Question 2: Find the explicit general solution of 2xyy’ – y^2 +x^2 = 0 (Homogeneous Substitution Method) 2xyy’ – y^2 + x^2 = 0 Or, dy/dx = (x^2 – y^2) By the substitution method, Let v = y/x Therefore, y = vx
DEFERENTIAL SOLUTIONS Or, dy/dx = v + x(dv/dx) Therefore, by substituting the value we get, v + x(dv/dx) = {x^2 – (v^2)(x^2)}/2vx^2 Or, dx/x = 2dv/((3v^2)-1) Integrating both sides we get, Logx + C1 = 1/3log ((3v^2)-1) + C2 Or, x^3= 3v^2 – 1 + C Where C is the composite constant Putting the value of v in the equation, we get, x^3= {3(y^2)/(x^2)} – 1 + C Question 3: Find the explicit general solution of xy’ + y + 4 =0.

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