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Consider the differential equation dy/dx=xy^4.

Find ⅆ2yⅆx2 in terms of x and y. Determine the concavity of all solution curves for the given differential equation in Quadrant II. Give a reason for your answer.

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Answer:

To find the second derivative of y with respect to x, we can take the derivative of the given differential equation with respect to x:

dy/dx = xy^4 d/dx(dy/dx) = d/dx(xy^4) d2y/dx2 = y^4 + 4xy^3(dy/dx)

Substituting the given differential equation into the expression for the second derivative, we get:

d2y/dx2 = y^4 + 4xy3(xy4) = y^4 + 4x2y7

In Quadrant II, x is negative and y is positive. Since y^4 and y^7 are both positive for any value of y, the expression for the second derivative simplifies to:

d2y/dx2 = y^4(1 + 4x2y3)

Since x is negative in Quadrant II, the quantity (1 + 4x2y3) is positive. Therefore, d2y/dx2 is positive for all values of x and y in Quadrant II. This means that all solution curves for the given differential equation are concave up in Quadrant II.

Received message. To find the second derivative of y with respect to x, we can take the derivative of the given differential equation with respect to x: dy/dx = xy^4 d/dx(dy/dx) = d/dx(xy^4) d2y/dx2 = y^4 + 4xy^3(dy/dx) Substituting the given differential equation into the expression for the second derivative, we get: d2y/dx2 = y^4 + 4xy^3(xy^4) = y^4 + 4x^2y^7 In Quadrant II, x is negative and y is positive. Since y^4 and y^7 are both positive for any value of y, the expression for the second derivative simplifies to: d2y/dx2 = y^4(1 + 4x^2y^3) Since x is negative in Quadrant II, the quantity (1 + 4x^2y^3) is positive. Therefore, d2y/dx2 is positive for all values of x and y in Quadrant II. This means that all solution curves for the given differential equation are concave up in Quadrant II.

Explanation:

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User Yaswant Narayan
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