Find the equation of the curve for which y^('')=(4)/(x^(3)) and which is tangent to the line 2x+y=5 at the point (1,3).

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Find the equation of the curve for which  y^('')=(4)/(x^(3)) and which is tangent to the line  2x+y=5 at the point  (1,3).

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Integrate to find derivative: Integrate the second derivative to find the first derivative. Given y'' = 4/x^3, we integrate to find y'. ∫(4/x^3)dx = ∫4x^(-3)dx y' = -2x^(-2) + C1 We need to find the constant C1.Find constant C1: Use the point of tangency to find the first constant C1. The curve is tangent to the line 2x + y = 5 at the point (1,3). This means that the derivative of the curve at x = 1 should be equal to the derivative of the line at that point. The derivative of the line 2x + y = 5 is -2 (since dy/dx = -2 when we differentiate 2x + y = 5 with respect to x). So, y'(1) = -2 should equal -2(1)^(-2) + C1. -2 = -2 + C1 C1 = 0Integrate for original function: Integrate the first derivative to find the original function y. Now that we have y' = -2x^(-2), we integrate to find y. ∫(-2x^(-2))dx = ∫-2x^(-2)dx y = 2x^(-1) + C2 We need to find the constant C2.Find constant C2: Use the point of tangency to find the second constant C2. The curve passes through the point (1,3), so we can use this to find C2. y(1) = 3 should equal 2(1)^(-1) + C2. 3 = 2 + C2 C2 = 1Write final curve equation: Write the final equation of the curve. Now that we have both constants, we can write the equation of the curve. y = 2x^(-1) + 1 Or, in a more conventional form: y = 2/x + 1

Find the equation of the curve for which $y^{\prime \prime}=\frac{4}{x^{3}}$ and which is tangent to the line $2 x+y=5$ at the point $(1,3)$ .

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Find the equation of the curve for which y′′=4x3 y^{\prime \prime}=\frac{4}{x^{3}} y′′=x34​ and which is tangent to the line 2x+y=5 2 x+y=5 2x+y=5 at the point (1,3) (1,3) (1,3).

Find the equation of the curve for which $y^{\prime \prime}=\frac{4}{x^{3}}$ and which is tangent to the line $2 x+y=5$ at the point $(1,3)$ .