NCERT/CBSE MATHEMATICS CLASS 12 textbook - TutorBreeze.com

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MISCELLANEOUS EXERCISES Answers to NCERT/CBSE MATH (Class XII)textbook Chapter 9 DIFFERENTIAL EQUATIONS

2.(i ) y = ae x + be− x + x 2 dy ⇒ = ae x − be− x + 2 x dx d2y ⇒ 2 = ae x + be− x + 2 dx y = ae x + be− x + x 2 ⇒ y − x 2 = ae x + be− x d2y = ae x + be− x + 2 = y − x 2 + 2 2 dx d2y dy Now x 2 + 2 − xy + x 2 − 2 dx dx 2 = x ( y − x + 2 ) + 2 ( ae x − be− x + 2 x ) − xy + x 2 − 2

⇒

= xy − x3 + 2 x + 2ae x − 2be− x + 4 x − xy + x 2 − 2 = − x3 + 2ae x − 2be − x + 6 x + x 2 − 2 ≠ 0 The given expression is not a solution of the given differential equation

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2.(ii y = e x (a cos x + b sin x) dy d de x = e x (a cos x + b sin x) + (a cos x + b sin x) dx dx dx dy ⇒ = e x (−a sin x + b cos x) + (a cos x + b sin x)e x dx x = e (−a sin x + b cos x + a cos x + b sin x) d2y de x x d =e (−a sin x + b cos x + a cos x + b sin x) + (−a sin x + b cos x + a cos x + b sin x) dx 2 dx dx 2 d y ⇒ 2 = e x (−a cos x − b sin x − a sin x + b cos x) + (−a sin x + b cos x + a cos x + b sin x)e x dx d2y ⇒ 2 = e x (−a cos x − b sin x − a sin x + b cos x − a sin x + b cos x + a cos x + b sin x) dx d2y ⇒ 2 = e x (−a sin x + b cos x − a sin x + b cos x) dx d2y ⇒ 2 = 2e x (−a sin x + b cos x) dx d2y dy ⇒ 2 − 2 + 2y dx dx x = 2e (−a sin x + b cos x ) − 2e x (−a sin x + b cos x + a cos x + b sin x) + 2e x (a cos x + b sin x) = 2e x (−a sin x + b cos x + a sin x − b cos x − a cos x − b sin x + a cos x + b sin x) = 2e x (0) = 0 The given expression is not a solution of the given differential equation




29(iii ) y = x sin 3 x dy d dx = x sin 3 x + sin 3 x dx dx dx dy ⇒ = 3 x cos 3 x + sin 3 x.1 dx d2y d dx d ⇒ 2 = 3[ x cos 3 x + cos 3 x ] + sin 3 x dx dx dx dx 2 d y ⇒ 2 = 3[ x(− sin 3 x) + cos 3 x.1] + 3cos 3 x dx d2y ⇒ 2 = −9 x sin 3 x + 3cos 3 x + 3cos 3 x dx d2y ⇒ 2 + 9 y − 6 cos 3 x = −9 x sin 3 x + 3cos 3 x + 3 cos 3 x + 9 x sin 3 x + 6 cos 3 x = 0 dx The given expression is not a solution of the given differential equation ⇒

2(iv) x 2 = 2 y 2 log y d 2 d 2 dy 2 x = 2 log y y + 2 log x dx dx dx 1 dy dy ⇒ 2x = 2 y2 + 2 log y (2 y ) y dx dx dy ⇒ x = y + log y (2 y ) dx dy ⇒ x = [ y + 2 y log y ] dx dy ⇒ xy =  y 2 + 2 y 2 log y  dx 2 2 But , x = 2 y log y ⇒ xy =  y 2 + x 2  ⇒

dy dx

dy 2  y + x 2  − xy = 0 dx 



