JEE Mathematics Differential Equations MCQs Set 06

Answer: (c) 2

Explanation:
1. Substituting \( y=e^{mx} \) gives the auxiliary equation \( m^3-3m^2-4m+12=0 \).
2. Factorise: \( (m-3)(m^2-4)=0 \), so \( m=3,2,-2 \).
3. Natural values are 2 and 3, so there are 2 of them.

Teacher's Note:
1. For a linear equation with constant coefficients, \( e^{mx} \) is a solution exactly when \( m \) is a root of the auxiliary equation.
2. Read the word "natural" carefully before counting roots.

Answer: (a) \( y=ce^{-\phi(x)}+\phi(x)-1 \)

Explanation:
1. The equation is linear: \( y'+\phi' y=\phi\phi' \), with integrating factor \( e^{\phi} \).
2. \( (ye^{\phi})'=\phi\phi' e^{\phi} \), so \( ye^{\phi}=\int \phi e^{\phi}\,d\phi=(\phi-1)e^{\phi}+c \).
3. Hence \( y=ce^{-\phi}+\phi-1 \).

Teacher's Note:
1. Spot the linear form \( y'+Py=Q \) with \( P=\phi' \) and use the integrating factor \( e^{\int P\,dx} \).
2. The integral \( \int te^{t}dt \) is done by parts.

Answer: (b) \( x^{4/3}-y^{4/3}=c \)

Explanation:
1. Differentiating the family: \( x^{-1/3}+y^{-1/3}y'=0 \), so \( y'=-(y/x)^{1/3} \).
2. For orthogonal trajectories replace \( y' \) by \( -1/y' \): \( y'=(x/y)^{1/3} \).
3. Separate: \( y^{1/3}dy=x^{1/3}dx \), so \( y^{4/3}-x^{4/3}=c' \), i.e. \( x^{4/3}-y^{4/3}=c \).

Teacher's Note:
1. Eliminate the parameter first, then replace \( dy/dx \) by \( -dx/dy \).
2. Orthogonal trajectories always lead to a new first-order equation that is solved by separating variables here.

Answer: (b) \( y=2\tan^{-1}(e^x-1)-x \)

Explanation:
1. Put \( v=x+y \): \( v'=1+\cos v+\sin v \).
2. With \( t=\tan(v/2) \): \( 1+\cos v+\sin v=\dfrac{2(1+t)}{1+t^2} \), and \( dv=\dfrac{2\,dt}{1+t^2} \), so \( \dfrac{dt}{1+t}=dx \).
3. Thus \( 1+t=ce^{x} \). At the origin \( v=0,t=0 \) gives \( c=1 \), so \( t=e^x-1 \).
4. \( x+y=2\tan^{-1}(e^x-1) \), i.e. \( y=2\tan^{-1}(e^x-1)-x \).

Teacher's Note:
1. The substitution \( v=x+y \) turns this into a variable-separable equation.
2. The half-angle substitution \( t=\tan(v/2) \) makes sine and cosine rational.

Answer: (b) \( 2xe^x+y^2e^y=0 \)

Explanation:
1. Note \( e^x(x+1)dx=d(xe^x) \). Let \( u=xe^x \): \( du-u\,dy+ye^y\,dy=0 \), so \( \dfrac{du}{dy}-u=-ye^y \).
2. Integrating factor \( e^{-y} \): \( (ue^{-y})'=-y \), so \( ue^{-y}=-\dfrac{y^2}{2}+C \).
3. With \( x=0,y=0 \) we get \( C=0 \), so \( 2xe^{x}+y^2e^{y}=0 \).

Teacher's Note:
1. Look for an exact differential such as \( d(xe^x) \) hidden in the equation.
2. Treating x as a function of y can make a difficult equation linear.

Answer: (c) 60 min

Explanation:
1. \( \dfrac{dy}{dt}=-k\sqrt y \), so \( \dfrac{dy}{\sqrt y}=-\dfrac{dt}{15} \).
2. Integrating: \( 2\sqrt y=-\dfrac{t}{15}+C \). At \( t=0,y=4 \): \( C=4 \).
3. Tank empty when \( y=0 \): \( 0=4-\dfrac{t}{15} \), so t = 60 min.

Teacher's Note:
1. Draining problems lead to separable equations with a negative sign for the falling level.
2. Use the initial depth to find the constant before putting y = 0.

Answer: (a) \( y=x\tan^{-1}\!\left(\ln\dfrac ex\right) \)

Explanation:
1. The equation is homogeneous: put \( y=vx \): \( v+xv'=v-\cos^2v \), so \( \sec^2v\,dv=-\dfrac{dx}{x} \).
2. Integrating: \( \tan v=-\ln x+c \). At \( x=1,v=\pi/4 \): \( c=1 \).
3. \( \tan\dfrac yx=1-\ln x=\ln\dfrac ex \), so \( y=x\tan^{-1}(\ln\frac ex) \).

Teacher's Note:
1. Whenever the right side depends only on \( y/x \), put \( y=vx \).
2. Use the given point to fix the constant.

Answer: (a) \( \dfrac\pi2<I<\dfrac{\pi^2}{4} \)

Explanation:
1. The left side is \( (f\sin x)' \), so \( f\sin x=x+c \). Boundedness at \( x\to0 \) forces \( c=0 \), hence \( f(x)=\dfrac{x}{\sin x} \).
2. For \( 0<x<\pi/2 \): \( 1<\dfrac{x}{\sin x}<\dfrac{\pi}{2} \) (increasing function). Therefore \( \dfrac\pi2<I<\dfrac\pi2\cdot\dfrac\pi2=\dfrac{\pi^2}{4} \).
3. (Numerically \( I\approx1.83 \).)

Teacher's Note:
1. Recognise \( f'\sin x+f\cos x \) as the derivative of \( f\sin x \).
2. The boundedness condition removes the constant that would blow up at 0.

Answer: (b) 2

Explanation:
1. Let the line be \( y=mx+c \), so \( y'=m \). Then \( m+xm^2-mx-c=0 \) for all x.
2. Comparing coefficients: \( m^2-m=0 \) and \( m-c=0 \).
3. So \( (m,c)=(0,0) \) or \( (1,1) \): the lines \( y=0 \) and \( y=x+1 \). There are 2 lines.

Teacher's Note:
1. For a straight line, substitute \( y=mx+c \) and equate coefficients of powers of x.
2. Each valid pair (m, c) gives exactly one line.

Answer: (a) \( e^{\pi} \)

Explanation:
1. This is a Bernoulli equation. Put \( u=\dfrac1y \): \( u'+u=-(\sin x+\cos x) \).
2. A particular solution is \( u=-\sin x \), so \( u=-\sin x+Ce^{-x} \).
3. \( u(0)=1 \) gives \( C=1 \). Then \( y(\pi)=\dfrac1{0+e^{-\pi}}=e^{\pi} \).

Teacher's Note:
1. For \( y'+Py=Qy^n \) divide by \( y^n \) and substitute \( u=y^{1-n} \).
2. Guess a particular solution of the form \( A\sin x+B\cos x \) when the right side has sine and cosine.

Answer: (c) m = 3/2

Explanation:
1. With \( y=u^m \): \( 2mx^4u^{2m-1}u'+u^{4m}=4x^6 \).
2. For homogeneity all terms must have the same degree, counting \( u' \) as degree 0: \( 4+2m-1=4m=6 \).
3. \( 4m=6 \) gives \( m=\dfrac32 \), and then \( 2m+3=6 \) as well.

Teacher's Note:
1. In a homogeneous equation every term has the same total degree in x and y.
2. Equate the degrees of all terms to find m.

Answer: (a) \( x^2+y^2=x^4 \)

Explanation:
1. The normal at \( (x,y) \) meets the x-axis at \( x+yy' \). So \( x(x+yy')=2(x^2+y^2) \), i.e. \( xyy'=x^2+2y^2 \).
2. With \( z=y^2 \): \( z'-\dfrac{4z}{x}=2x \), so \( (zx^{-4})'=2x^{-3} \) and \( zx^{-4}=-x^{-2}+c \).
3. Through (1, 0): \( c=1 \), so \( y^2=x^4-x^2 \), i.e. \( x^2+y^2=x^4 \).

Teacher's Note:
1. The intercept of the normal on the x-axis is \( x+yy' \).
2. The substitution \( z=y^2 \) turns the equation into a linear one.

Answer: (c) approximately 10 hours

Explanation:
1. \( \dfrac{dM}{dt}=-kM \), so \( M=M_0e^{-kt} \). Half is lost in 1 hour, so \( e^{-k}=\dfrac12 \).
2. Losing 99.9% leaves 0.001: \( 2^{-t}=0.001 \).
3. \( t=\log_2 1000\approx9.97 \) hours, which is approximately 10 hours (slightly less than 10).

Teacher's Note:
1. Rate proportional to the amount present means exponential decay.
2. Since \( 2^{10}=1024\approx1000 \), the time is just under 10 hours.

Answer: (d) \( -\dfrac{y^2}{x^2} \)

Explanation:
1. Let \( L=\ln|cx| \), so \( y=\dfrac xL \) and \( L'=\dfrac1x \).
2. \( y'=\dfrac1L-\dfrac{xL'}{L^2}=\dfrac1L-\dfrac1{L^2} \).
3. Since \( \dfrac yx=\dfrac1L \), we get \( y'=\dfrac yx-\left(\dfrac yx\right)^2 \).
4. So \( \phi\!\left(\dfrac xy\right)=-\dfrac{y^2}{x^2} \).

Teacher's Note:
1. Differentiate the given solution and compare it with the form of the differential equation.
2. The constant c disappears automatically after differentiation.

Answer: (a) \( x+a\ln\left(\sqrt{y^2-a^2}+y\right)=c \)

Explanation:
1. The normal makes an angle \( \theta \) with the ordinate where \( \tan\theta=y' \). The projection of the ordinate on the normal is \( y\cos\theta=\dfrac{y}{\sqrt{1+y'^2}}=a \).
2. So \( y'=\pm\dfrac{\sqrt{y^2-a^2}}{a} \).
3. Taking the negative sign: \( \dfrac{dy}{\sqrt{y^2-a^2}}=-\dfrac{dx}{a} \), giving \( a\ln\left(y+\sqrt{y^2-a^2}\right)=-x+c \).
4. Hence \( x+a\ln\left(\sqrt{y^2-a^2}+y\right)=c \).

Teacher's Note:
1. The projection of the ordinate on the normal is \( y\cos\theta \), where \( \theta \) is the angle the tangent makes with the x-axis.
2. The other sign of the slope gives the mirror-image family.

Answer: (a) \( y=2-(2+a^2)e^{\frac{x^2-a^2}{2}} \)

Explanation:
1. Differentiate: \( xy=2x+y' \), so \( y'-xy=-2x \).
2. \( \left(ye^{-x^2/2}\right)'=-2xe^{-x^2/2} \), so \( y=2+Ce^{x^2/2} \).
3. Putting \( x=a \) in the original equation: \( 0=a^2+y(a) \), so \( 2+Ce^{a^2/2}=-a^2 \).
4. \( C=-(2+a^2)e^{-a^2/2} \), giving \( y=2-(2+a^2)e^{\frac{x^2-a^2}{2}} \).

Teacher's Note:
1. Differentiating an integral equation gives a differential equation, but the original equation supplies an extra condition.
2. Here the condition is that the integral vanishes at \( x=a \).

Answer: (a) \( f(x)=c\,x^{\frac{1-n}{n}} \)

Explanation:
1. Put \( u=tx \): \( \dfrac1x\int_0^xf(u)\,du=nf(x) \). Let \( F(x)=\int_0^xf \).
2. Then \( F=nxF' \), so \( \dfrac{F'}{F}=\dfrac1{nx} \) and \( F=cx^{1/n} \).
3. \( f=F'=\dfrac cn x^{\frac1n-1}=c'\,x^{\frac{1-n}{n}} \).

Teacher's Note:
1. Substituting \( u=tx \) turns the integral into a standard indefinite-type integral of f.
2. Differentiating the resulting relation gives a separable equation.

Answer: (d) \( xy=6 \)

Explanation:
1. Differentiate: \( xy=2xy+x^2y' \), so \( x^2y'=-xy \).
2. \( \dfrac{y'}{y}=-\dfrac1x \), so \( xy=c \).
3. Through (2, 3): \( c=6 \), so xy = 6.

Teacher's Note:
1. Differentiate both sides using the Leibniz rule for the upper limit.
2. The curve is a rectangular hyperbola.

Answer: (b) \( x^2+y^3+y\cos2x=C \)

Explanation:
1. \( M=2x-2y\sin2x,\ N=3y^2+\cos2x \). \( M_y=-2\sin2x=N_x \), so the equation is exact.
2. \( \int M\,dx=x^2+y\cos2x \) (treating y as constant).
3. Add the terms of N free of x: \( y^3 \). Solution: \( x^2+y^3+y\cos2x=C \).

Teacher's Note:
1. Always test \( M_y=N_x \) before choosing a method.
2. For an exact equation, integrate M with respect to x and add the missing y-only terms from N.

Answer: (c) \( xy^2=2y^5+C \)

Explanation:
1. Write as \( \dfrac{dx}{dy}+\dfrac{2x}{y}=10y^2 \) (linear in x).
2. Integrating factor \( e^{\int 2/y\,dy}=y^2 \): \( (xy^2)'=10y^4 \).
3. \( xy^2=2y^5+C \).

Teacher's Note:
1. When the equation is not linear in y, try treating x as the dependent variable.
2. The integrating factor here is \( y^2 \).

Answer: (c) 4

Explanation:
1. \( \operatorname{tr}(AB)=x^2y^{2/3}+\left(x\,y'\right)^{1/3} \). The equation is \( y'-x^2y^{2/3}=(xy')^{1/3} \).
2. Cube both sides to remove the fractional power of \( y' \): \( \left(y'-x^2y^{2/3}\right)^3=xy' \).
3. The highest derivative is \( y' \), so order \( m=1 \), and its highest power is 3, so degree \( n=3 \).
4. \( m+n=4 \).

Teacher's Note:
1. Degree is defined only after the equation is made polynomial in the derivatives.
2. Powers of y itself do not affect the degree.

Answer: (d) 1

Explanation:
1. \( \left(f e^{2x}\right)'\le e^{2x} \), so \( f(x)e^{2x}\le\dfrac{e^{2x}-1}{2} \).
2. Hence \( f(x)\le\dfrac{1-e^{-2x}}{2}<\dfrac12 \) for all \( x\ge0 \).
3. So the value 1 is impossible, while \( \dfrac14 \), 0 and \( -3 \) can all be attained (for example by \( f=\frac{1-e^{-2x}}{2} \) and \( f=-100x \)).

Teacher's Note:
1. Multiply by the integrating factor \( e^{2x} \) to turn the inequality into a statement about a derivative.
2. Integrating an inequality between 0 and x keeps its direction.

Answer: (a) I, II and IV only

Explanation:
1. Multiply by \( y' \): \( x(y')^2+3=y^2y' \).
2. Highest derivative is \( y' \), so the order is 1; its highest power is 2, so the degree is 2.
3. The term \( (y')^2 \) (and \( y^2 \)) makes it non-linear. Hence I, II and IV are correct.

Teacher's Note:
1. Clear fractions in the derivative before reading off the degree.
2. A linear equation has y and its derivatives only to the first power.

Answer: (a) I, II and III only

Explanation:
1. (I) degree \( -1 \): homogeneous. (II) degree \( \tfrac13-\tfrac23=-\tfrac13 \): homogeneous.
2. (III) \( x\ln\dfrac{\sqrt{x^2+y^2}}{y} \) and \( ye^{x/y} \) are both of degree 1: homogeneous.
3. (IV) The first part \( x\ln\dfrac{2x^2+y^2}{x(x+y)} \) has degree 1, but \( y^2\tan(\cdot) \) has degree 2. Mixed degrees, so not homogeneous.

Teacher's Note:
1. A function is homogeneous if \( f(\lambda x,\lambda y)=\lambda^nf(x,y) \).
2. All terms must have the same degree n.

Answer: (b) 3 or \( -3 \)

Explanation:
1. At \( x=0 \): \( (f'(0))^2=f(0)f''(0)=1\cdot9=9 \).
2. So \( f'(0)=\pm3 \).
3. Both occur: \( f=e^{3x} \) and \( f=e^{-3x} \) satisfy \( (f')^2=ff'' \) with \( f(0)=1,\ f''(0)=9 \).

Teacher's Note:
1. Substitute the initial values directly into the differential equation.
2. Always check that each root can actually be attained by a function.

Answer: (a) A-R, B-S, C-P, D-Q

Explanation:
1. (A) \( \dfrac{dV}{dt}=-k\cdot4\pi r^2 \) with \( V=\dfrac43\pi r^3 \) gives \( \dfrac{dr}{dt}=-k \), so \( \lambda=1 \): R.
2. (B) \( y'''-13y'=12e^{4x}+24e^{-x}=12y \), so \( K=12 \): S.
3. (C) \( \sin kt \): \( k^2=9 \); \( e^{kt} \): \( k^2+k-6=0\Rightarrow k=2,-3 \). Common value \( k=-3 \): P.
4. (D) Homogeneity needs \( 3\alpha+1=\alpha-1 \), so \( \alpha=-1 \): Q.

Teacher's Note:
1. For matching questions solve each item independently, then compare with the options.
2. Each of the four answers in Column II is used exactly once.

Answer: (b) 2

Explanation:
1. \( y'-\dfrac{2x}{1+x^2}y=x^2-1 \). Integrating factor \( \dfrac1{1+x^2} \).
2. \( \left(\dfrac y{1+x^2}\right)'=\dfrac{x^2-1}{1+x^2}=1-\dfrac2{1+x^2} \).
3. \( \dfrac y{1+x^2}=x-2\tan^{-1}x+c \), and \( y(0)=0 \) gives \( c=0 \).
4. \( f(-1)=2\left(-1+\dfrac\pi2\right)=\pi-2\approx1.14 \). The least integer not less than this is 2.

Teacher's Note:
1. A linear equation with a rational coefficient has an integrating factor that is a rational function.
2. The "least integer greater than or equal to" is the ceiling of the value.