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MATH150T3 - MATH 150 Linear Equations and its Application...

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MATH 150 Linear Equations and it’s Application on Compound interest Cheung Man Wai Tutorial 3 1 Linear Equation A first-order linear differential equation appears in the form dy dx + p ( x ) y = g ( x ) , with y ( x 0 ) = y 0 . Applying the so-called integrating factor, the solution to (2) is given by y = 1 μ ( x ) y 0 + x x 0 μ ( x ) g ( x ) dx , where μ ( x ) = exp x x 0 p ( x ) dx . Above formulas are remarkably and explicitly applicable, which enable us to perform practical solving directly. Example 1 Find the solution of ty + 2 y = sin t, y ( π/ 2) = 1 , t > 0 . First note that this equation is not separable. Divide the equation by t to obtain the standard form y + 2 t y = sin t t , which gives us p ( t ) = 2 /t and g ( t ) = (sin t ) /t. Thus, μ ( t ) = exp t π/ 2 2 t dt = 4 π 2 t 2 1
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and y = π 2 4 t 2 1 + t π/ 2 4 π 2 t 2 sin t t dt = π 2 4 t 2 1 + 4 π 2 ( - t cos t + sin t ) t π/ 2 = 1 t 2 π 2 4 - 1 - t cos t + sin t . Example 2 Solve the initial value problem y - 1 2 y = 2 cos t, y (0) = a and describe the behavior of the solution corresponding to the initial value a.
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