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Vector Algebra

Question
CBSEENMA12032933
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Form the differential equation representing the family of curves y = a cos (x + b) where a and b are arbitrary constants.

Solution

The given equation is straight y space equals space straight a space cos left parenthesis straight x plus straight b right parenthesis               ...(1)
Differentiating w.r.t.x, we get,   dy over dx equals negative straight a space sin left parenthesis straight x plus straight b right parenthesis
Again differentiating w.r.t. x, we get, 
                     fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus straight a space cos space left parenthesis straight x plus straight b right parenthesis space space space space space space or space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative straight y space space space space space space space space space space left square bracket because space space of space left parenthesis 1 right parenthesis right square bracket
or              fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight y space equals space 0
which is required differential equation.

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = ex

Determine order and degree (if defined) of differential equations:
y'' + (y')2 + 2 y = 0

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