y′′−2y′+5y=0y″−2y′+5y=0
and a solution y1=e^xsin(2x)
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Given the problem y′′−2y′+5y=0y″−2y′+5y=0 and a solution y1=e^xsin(2x)
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Given the problem y′′−2y′+5y=0y″−2y′+5y=0 and a solution y1=e^xsin(2x)
Given the problem
y′′−2y′+5y=0y″−2y′+5y=0
and a solution y1=e^xsin(2x)
(1 point) Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + ap(x) y = 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions 3₁, 32. But there are times when only one function, call it y₁, is available and we would like to find a second linearly independent solution. We can find 2 using the method of reduction of order. First, under the necessary assumption the a₂ (2) #0 we rewrite the equation as Then the method of reduction of order gives a second linearly independent solution as and a solution 3/1 e² sin(2x) Applying the reduction of order method we obtain the following So we have y" + p(x)y' +q(x)y = 0 p(x) = p(x) = So the general solution to y" - 2y + 4y = 0 can be written as - Jp(z)dz y(x) where C' is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful cholce is to choose C so that all the constants in front reduce to 1. For example, if we obtain y2 = C3e² then we can choose C = 1/3 so that y/₂ = ²². Given the problem 32(x) = Cy₁u = Cy₁(z) | -dx a₁(x) a₂(x) y (1) = -S y = C₁Y/1 + C₂3/2 = C₁ y" - 2y + 5y = 0 and e-p(2)dx= -Sp(z)dz y(x) Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at y₂(x) = Cy₁u = q(z) = dx = +0₂ ao(x) a₂(z)' da
y′′−2y′+5y=0y″−2y′+5y=0
and a solution y1=e^xsin(2x)
(1 point) Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + ap(x) y = 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions 3₁, 32. But there are times when only one function, call it y₁, is available and we would like to find a second linearly independent solution. We can find 2 using the method of reduction of order. First, under the necessary assumption the a₂ (2) #0 we rewrite the equation as Then the method of reduction of order gives a second linearly independent solution as and a solution 3/1 e² sin(2x) Applying the reduction of order method we obtain the following So we have y" + p(x)y' +q(x)y = 0 p(x) = p(x) = So the general solution to y" - 2y + 4y = 0 can be written as - Jp(z)dz y(x) where C' is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful cholce is to choose C so that all the constants in front reduce to 1. For example, if we obtain y2 = C3e² then we can choose C = 1/3 so that y/₂ = ²². Given the problem 32(x) = Cy₁u = Cy₁(z) | -dx a₁(x) a₂(x) y (1) = -S y = C₁Y/1 + C₂3/2 = C₁ y" - 2y + 5y = 0 and e-p(2)dx= -Sp(z)dz y(x) Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at y₂(x) = Cy₁u = q(z) = dx = +0₂ ao(x) a₂(z)' da