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NCERT Solutions Class 12 Maths (Determinants) Miscellaneous Exercise

NCERT Solutions Class 12 Maths (Determinants) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Miscellaneous Exercise  This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 Maths (Determinants) Miscellaneous Exercise

Q1. 

Answer.  (i) The differential equation is given as: d2ydx2+5x(dydx)26y=logxd2ydx2+5x(dydx)26ylogx=0 The highest order derivative present in the differential equation is one.  The highest power raised to d2ydx2 is one. Hence, its degree is one.   (ii) The differential equation is given as: (dydx)34(dydx)2+7y=sinx(dydx)34(dydx)2+7ysinx=0 The highest order derivative present in the differential equation is dydx. Thus, its order is one.  The highest power raised to dydx is three. Hence, its degree is three.  

Q2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. (i) y=aex+bex+x2:xd2ydx2+2dydxxy+x22=0(ii) y=ex(acosx+bsinx):d2ydx22dydx+2y=0 

Answer.  (i) y=aex+bex+x2 Differentiating both sides with respect to x, we get: dydx=addx(ex)+bddx(ex)+ddx(x2)dydx=aexbex+2x Again, differentiating both sides with respect to x, we get: d2ydx2=aex+bex+2  Now, on substituting the values of dydx and d2ydx2 in the differential equation, we get:  L.H.S. xd2ydx2+2dydxxy+x22 =x(aex+bex+2)+2(aexbex+2x)x(aex+bex+x2)+x22=(axex+bxex+2x)+(2aex2bex+4x)(axex+bxex+x3)+x22=2aex2bex+x2+6x20 ⇒ L.H.S. ≠ R.H.S. Hence, the given function is not a solution of the corresponding differential equation. (ii) y=ex(acosx+bsinx)=aexcosx+bexsinx Differentiating both sides with respect to x, we get: dydx=addx(excosx)+bddx(exsinx)dydx=a(excosxexsinx)+b(exsinx+excosx)dydx=(a+b)excosx+(ba)exsinxAgain, differentiating both sides with respect to x, we get:d2ydx2=(a+b)ddx(excosx)+(ba)ddx(exsinx) d2ydx2=(a+b)[excosxexsinx]+(ba)[exsinx+excosx]d2ydx2=ex[(a+b)(cosxsinx)+(ba)(sinx+cosx)]d2ydx2=ex[acosxasinx+bcosxbsinx+bsinx+bcosxasinxacosx]d2ydx2=[2ex(bcosxasinx)]  Now, on substituting the values of d2ydx2 and dydx in the L.H.S. of the given differential  equation, we get: d2ydx2+2dydx+2y =2ex(bcosxasinx)2ex[(a+b)cosx+(ba)sinx]+2ex(acosx+bsinx)=ex[(2bsinx2asinx)(2acosx+2bcosx)=ex[(2bsinx2asinx)+(2acosx+2bsinx)]=ex[(2b2a2b+2a)cosx]+ex[(2a2b+2a+2b)sinx]=0 Hence, the given function is a solution of the corresponding differential equation. (iii) y=xsin3x Differentiating both sides with respect to x, we get: dydx=ddx(xsin3x)=sin3x+xcos3x3dydx=sin3x+3xcos3x Again, differentiating both sides with respect to x, we get: d2ydx2=ddx(sin3x)+3ddx(xcos3x) d2ydx2=3cos3x+3[cos3x+x(sin3x)3]d2ydx2=6cos3x9xsin3x Substituting the value of d2ydx2 in the L.H.S. of the given differential equation, we get: d2ydx2+9y6cos3x =(6cos3x9xsin3x)+9xsin3x6cos3x=0 Hence, the given function is a solution of the corresponding differential equation.   (iv) x2=2y2logy Differentiating both sides with respect to x, we get: 2x=2ddx=[y2logy]x=[2ylogydydx+y21ydydx]x=dydx(2ylogy+y)dydx=xy(1+2logy)  Substituting the value of dydx in the L.H.S. of the given differential equation, we get: (x2+y2)dydxxy 

Q3.