Question 83 of 96intermediate🔧 ApplyShort Answer6 marks

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i)

xy=aex+be−x+x2xy = a e^x + b e^{-x} + x^2 : xd2ydx2+2dydx−xy+x2−2=0x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0

Answer

Differentiating xy = aexae^{x} + be^(-x) + x² twice and substituting in the differential equation gives L.H.S. = 0 = R.H.S. Hence verified.

(ii)

y=ex(acos⁡x+bsin⁡x)y = e^x (a \cos x + b \sin x) : d2ydx2−2dydx+2y=0\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0

Answer

Differentiating y = exe^{x}(a cos x + b sin x) twice and substituting in the differential equation gives L.H.S. = 0 = R.H.S. Hence verified.

(iii)

y=xsin⁡3xy = x \sin 3x : d2ydx2+9y−6cos⁡3x=0\frac{d^2 y}{dx^2} + 9y - 6\cos 3x = 0

Answer

Differentiating y = x sin 3x twice and substituting in the differential equation gives L.H.S. = 0 = R.H.S. Hence verified.

(iv)

x2=2y2log⁡yx^2 = 2y^2 \log y : (x2+y2)dydx−xy=0(x^2 + y^2) \frac{dy}{dx} - xy = 0

Answer

Differentiating x² = 2y² log y implicitly and substituting in the differential equation gives L.H.S. = 0 = R.H.S. Hence verified.

Explanation

This question requires verifying that given functions are solutions of corresponding differential equations. The method involves differentiating the given function (explicitly or implicitly) the required number of times, then substituting these derivatives into the differential equation to verify that L.H.S. equals R.H.S.

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