JEE Mathematics Differential Equations MCQs Set 07

Answer: (a) \( \frac{x\ln x+y\ln y}{xy}=c \)

Explanation:
1. Write the proposed solution as \( F=\frac{\ln x}{y}+\frac{\ln y}{x}=c \) (this equals (a)).
2. Then \( dF=\left(\frac1{xy}-\frac{\ln y}{x^2}\right)dx+\left(\frac1{xy}-\frac{\ln x}{y^2}\right)dy=0 \).
3. \( \frac{dy}{dx}=\frac{(x-y\ln y)/(x^2y)}{(x\ln x-y)/(xy^2)}=\frac{y(x-y\ln y)}{x(x\ln x-y)} \), which is the given equation.

Teacher's Note:
1. Test an option by differentiating it implicitly and comparing with the given equation.
2. Simplify (x ln x + y ln y)/(xy) as ln x / y + ln y / x.

Answer: (c) \( \left|\frac{\sin x}{xy}\right|=c \)

Explanation:
1. \( d\left(\frac{\sin x}x\right)=\frac{x\cos x-\sin x}{x^2}dx \), so the left side is \( x^2\,d\!\left(\frac{\sin x}{x}\right) \).
2. The equation becomes \( \frac{d(\sin x/x)}{\sin x/x}=\frac{dy}{y} \).
3. Integrating: \( \ln\left|\frac{\sin x}x\right|=\ln|y|+k \), i.e. \( \left|\frac{\sin x}{xy}\right|=c \).

Teacher's Note:
1. Recognise the quotient-rule derivative of sin x / x.
2. Separate the variables once the left side is written in terms of sin x / x.

Answer: (c) \( \frac1{(x+y)^2}=x^2+1+ce^{x^2} \)

Explanation:
1. Let \( v=x+y \): \( v'=1+y'=x^3v^3-xv \), a Bernoulli equation.
2. Put \( w=\frac1{v^2} \): \( w'=-\frac{2v'}{v^3}=-2x^3+2xw \), i.e. \( w'-2xw=-2x^3 \).
3. Integrating factor \( e^{-x^2} \): \( \left(we^{-x^2}\right)'=-2x^3e^{-x^2} \), so \( we^{-x^2}=(x^2+1)e^{-x^2}+c \), giving \( \frac1{(x+y)^2}=x^2+1+ce^{x^2} \).

Teacher's Note:
1. Substituting v = x + y reduces the equation to a Bernoulli equation.
2. Then w = 1/v² makes it linear.

Answer: (a) \( f(x)=\frac{2x}{2-x^2} \)

Explanation:
1. \( \frac{dy}{dx}=\frac yx+y^2 \). Put \( u=\frac1y \): \( u'=-\frac{y'}{y^2}=-\frac ux-1 \), so \( (xu)'=-x \).
2. \( xu=-\frac{x^2}2+C \), i.e. \( \frac xy=C-\frac{x^2}2 \). The point (1, 2): \( \frac12=C-\frac12 \), so C = 1.
3. Then \( \frac xy=\frac{2-x^2}{2} \) and \( y=\frac{2x}{2-x^2} \).

Teacher's Note:
1. This is a Bernoulli equation in y; use u = 1/y.
2. The integrating factor of u′ + u/x = −1 is x.

Answer: (a) \( y^2=\sin x \)

Explanation:
1. Put \( v=y^2 \): \( v'\sin x+v\cos x=2\sin x\cos x \), i.e. \( (v\sin x)'=\sin2x \).
2. Integrating: \( v\sin x=\sin^2x+C \). At (π/2, 1): \( 1=1+C\Rightarrow C=0 \).
3. So \( y^2=\sin x \).

Teacher's Note:
1. Substituting v = y² turns the equation into a linear one.
2. The left side becomes (v sin x)′.

Answer: (a) \( x^2\cos y+y^2\sin x=c \)

Explanation:
1. \( \frac{\partial M}{\partial y}=-2x\sin y+2y\cos x=\frac{\partial N}{\partial x} \): the equation is exact.
2. Integrate M with respect to x: \( x^2\cos y+y^2\sin x \) (this already accounts for all terms of N).
3. The solution is \( x^2\cos y+y^2\sin x=c \).

Teacher's Note:
1. Check exactness by comparing ∂M/∂y and ∂N/∂x.
2. Integrate M with respect to x and fix the function of y using N.

Answer: (b) \( \sec y=x+1+ce^x \)

Explanation:
1. Divide by \( \cos^2y \): \( \sec y\tan y\,y'=\sec y-x \). Put \( u=\sec y \): \( u'=u-x \).
2. Linear equation: \( (ue^{-x})'=-xe^{-x} \), so \( ue^{-x}=(x+1)e^{-x}+c \).
3. Hence \( \sec y=x+1+ce^x \).

Teacher's Note:
1. Dividing by cos² y makes sec y the natural variable.
2. Solve the linear equation u′ − u = −x with integrating factor e^(−x).

Answer: (b) \( [1+(y')^2]^3=a^2(y'')^2 \)

Explanation:
1. Differentiate twice to eliminate h and k: \( (x-h)+(y-k)y'=0 \) and \( 1+(y')^2+(y-k)y''=0 \).
2. So \( y-k=-\frac{1+y'^2}{y''} \) and \( x-h=\frac{(1+y'^2)y'}{y''} \).
3. Substitute: \( \frac{(1+y'^2)^2(y'^2+1)}{y''^2}=a^2\Rightarrow[1+(y')^2]^3=a^2(y'')^2 \).

Teacher's Note:
1. Two arbitrary constants h and k require two differentiations.
2. This is the radius-of-curvature condition.

Answer: (b) \( y_1+y_1^3-xy_2=0 \)

Explanation:
1. The family is \( x^2+(y-k)^2=r^2 \) with two constants k and r.
2. Differentiate: \( x+(y-k)y_1=0 \) and again \( 1+y_1^2+(y-k)y_2=0 \).
3. From the first, \( y-k=-\frac{x}{y_1} \). Substituting: \( 1+y_1^2-\frac{xy_2}{y_1}=0 \), i.e. \( y_1+y_1^3-xy_2=0 \).

Teacher's Note:
1. Centres on the y-axis means the centre is (0, k).
2. Eliminate (y − k) between the two differentiated equations.

Answer: (b) \( \log_{4/3}2 \)

Explanation:
1. \( x=2e^{-at},\ y=e^{-bt} \).
2. \( \frac{x(1)}{y(1)}=2e^{b-a}=\frac32\Rightarrow e^{b-a}=\frac34 \).
3. x = y gives \( 2e^{-at}=e^{-bt}\Rightarrow e^{(b-a)t}=\frac12\Rightarrow\left(\frac34\right)^t=\frac12 \), so \( t=\frac{\ln2}{\ln(4/3)}=\log_{4/3}2 \).

Teacher's Note:
1. Solve each equation as an exponential decay.
2. Only the difference b − a matters.

Answer: (d) \( -2\tan^{-1}\left(\frac\pi2-1\right) \)

Explanation:
1. Let \( v=x+y \): \( v'=1+\cos v=2\cos^2\frac v2 \), so \( \frac12\sec^2\frac v2\,dv=dx \) and \( \tan\frac v2=x+C \).
2. At x = π/2, v = π/2: \( 1=\frac\pi2+C\Rightarrow C=1-\frac\pi2 \).
3. At x = 0, v = y(0): \( \tan\frac{y(0)}2=1-\frac\pi2 \), so \( y(0)=2\tan^{-1}\left(1-\frac\pi2\right)=-2\tan^{-1}\left(\frac\pi2-1\right) \).

Teacher's Note:
1. Substitute v = x + y to separate the variables.
2. 1 + cos v = 2cos²(v/2).

Answer: (d) \( e^4 \)

Explanation:
1. Linear form: \( y'-\frac2xy=2x^3+x \). The integrating factor is \( x^{-2} \).
2. \( \left(\frac y{x^2}\right)'=2x+\frac1x\Rightarrow\frac y{x^2}=x^2+\ln x+C \).
3. y(1) = 0 gives C = −1, so \( y=x^2(x^2+\ln x-1) \) and \( y(e)=e^2\cdot e^2=e^4 \).

Teacher's Note:
1. Rewrite the equation as a linear equation in y.
2. Use the initial condition to fix the constant.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The general circle \( x^2+y^2+2gx+2fy+c=0 \) has three independent constants, so its differential equation has order 3: A is true.
2. R is true: three non-collinear points fix g, f and c uniquely, which is why exactly three constants are needed. R explains A.

Teacher's Note:
1. The order of the differential equation equals the number of independent arbitrary constants.
2. Three constants correspond to three conditions needed to fix a circle.