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Exercise:
Convert the following systems of differential s o a single higher-order differential . abcliste abc dot x_ - x_+x_ dot x_ x_ - x_ abc dot x_ x_ + x_ dot x_ - x_ + x_ dot x_ x_ - x_ abcliste

Solution:
abcliste abc We can express the second derivative of x_ as follows using the differential for x_: ddot x_ -dot x_+dot x_ -dot x_+x_-x_ labelx The differential for x_ can be solved for x_: x_ dot x_+x_ Substituting x_ in refx leads to ddot x_ -dot x_+x_-dot x_+x_ -dot x_+x_-dot x_-x_ result-dot x_-x_ This is a second-order differential for x_. vspace.cm Alternatively we could also derive a second-order differential for x_: ddot x_ dot x_-dot x_ - x_+x_-dot x_ - x_+x_-dot x_ -dot x_+ x_+x_-dot x_ - dot x_-x_ This is the same differential as before. abc Deriving the differential for x_ and plugging in the expressions for dot x_ and dot x_ yields ddot x_ dot x_ + dot x_ x_ + x_ - x_ + x_ x_ + x_ Deriving again we find dddot x_ dot x_ + dot x_ x_ + x_ + x_ - x_ x_ + x_ - x_ We can express the term x_ - x_ in terms of dot x_ and ddot x_: dot x_-ddot x_ x_ + x_ - x_ - x_ x_ - x_ Upon substituting this expression in the for dddot x_ we find dddot x_ result-ddot x_ + dot x_ + x_ abcliste
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Exercise:
Convert the following systems of differential s o a single higher-order differential . abcliste abc dot x_ - x_+x_ dot x_ x_ - x_ abc dot x_ x_ + x_ dot x_ - x_ + x_ dot x_ x_ - x_ abcliste

Solution:
abcliste abc We can express the second derivative of x_ as follows using the differential for x_: ddot x_ -dot x_+dot x_ -dot x_+x_-x_ labelx The differential for x_ can be solved for x_: x_ dot x_+x_ Substituting x_ in refx leads to ddot x_ -dot x_+x_-dot x_+x_ -dot x_+x_-dot x_-x_ result-dot x_-x_ This is a second-order differential for x_. vspace.cm Alternatively we could also derive a second-order differential for x_: ddot x_ dot x_-dot x_ - x_+x_-dot x_ - x_+x_-dot x_ -dot x_+ x_+x_-dot x_ - dot x_-x_ This is the same differential as before. abc Deriving the differential for x_ and plugging in the expressions for dot x_ and dot x_ yields ddot x_ dot x_ + dot x_ x_ + x_ - x_ + x_ x_ + x_ Deriving again we find dddot x_ dot x_ + dot x_ x_ + x_ + x_ - x_ x_ + x_ - x_ We can express the term x_ - x_ in terms of dot x_ and ddot x_: dot x_-ddot x_ x_ + x_ - x_ - x_ x_ - x_ Upon substituting this expression in the for dddot x_ we find dddot x_ result-ddot x_ + dot x_ + x_ abcliste
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Differential equations
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second-order
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