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Exercise 1 · Q5

Q.Determine the order and degree (if defined) of the differential equation: (y′′′)2+(y′′)3+(y′)4+y5=0(y''')^2+(y'')^3+(y')^4+y^5=0, where y′=dydxy'=\frac{dy}{dx}, y′′=d2ydx2y''=\frac{d^2y}{dx^2} and y′′′=d3ydx3y'''=\frac{d^3y}{dx^3}

Yanam CbseNCERTSubjective· 2mImportance★★★★★
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The highest derivative is y′′′y''' (third order) and it is raised to the power 22, so order =3=3 and degree =2=2.

Order = order of the highest derivative present. Degree = power of the highest-order derivative when the equation is polynomial in derivatives.

Given: (y′′′)2+(y′′)3+(y′)4+y5=0(y''')^{2}+(y'')^{3}+(y')^{4}+y^{5}=0, where y′=dydxy'=\dfrac{dy}{dx}, y′′=d2ydx2y''=\dfrac{d^2y}{dx^2}, y′′′=d3ydx3y'''=\dfrac{d^3y}{dx^3}.

  1. Highest derivative present is y′′′=d3ydx3y'''=\dfrac{d^3y}{dx^3} → order =3=3. …

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