JEE Mathematics Differential Equations MCQs Set 24

Answer: (c) \( x\log_e|y|=2(x-1) \)

Explanation:
1. Separate the variables: \( \frac{dy}y=\frac{2}{x^2}dx \), so \( \log_e|y|=-\frac2x+C \).
2. The centre of the circle is \( (1,1) \), so \( 0=-2+C\Rightarrow C=2 \).
3. \( \log_e|y|=2-\frac2x \), i.e. \( x\log_e|y|=2(x-1) \). Answer: (c)

Teacher's Note:
1. Separate the variables and integrate.
2. Find the centre of the circle by completing squares.

Answer: (d) \( e-2 \)

Explanation:
1. \( \frac{dy}{dx}+(\sec^2x)y=\sec^2x\tan x \) is linear with integrating factor \( e^{\tan x} \). Put \( t=\tan x \): \( ye^{t}=\int te^tdt=e^t(t-1)+C \).
2. So \( y=\tan x-1+Ce^{-\tan x} \). \( y(0)=0\Rightarrow-1+C=0\Rightarrow C=1 \).
3. At \( x=-\frac\pi4 \): \( y=-1-1+e=e-2 \). Answer: (d)

Teacher's Note:
1. Use the integrating factor \( e^{\\tan x} \).
2. Integrate by parts after substituting \( t=\tan x \).

Answer: (a) \( y'\left(\frac\pi4\right)-y'\left(-\frac\pi4\right)=\pi-\sqrt2 \)

Explanation:
1. The integrating factor is \( e^{\int\tan x\,dx}=\sec x \). Then \( y\sec x=\int(2x+x^2\tan x)\sec x\,dx \), and since \( \int x^2\sec x\tan x\,dx=x^2\sec x-\int2x\sec x\,dx \), we get \( y\sec x=x^2\sec x+C \).
2. \( y(0)=1\Rightarrow C=1 \), so \( y=x^2+\cos x \) and \( y'=2x-\sin x \).
3. \( y'\left(\frac\pi4\right)-y'\left(-\frac\pi4\right)=2\left(\frac\pi2-\frac1{\sqrt2}\right)=\pi-\sqrt2 \). The other three statements fail: \( y'(\frac\pi4)+y'(-\frac\pi4)=0 \), \( y(\frac\pi4)+y(-\frac\pi4)=\frac{\pi^2}8+\sqrt2 \), \( y(\frac\pi4)-y(-\frac\pi4)=0 \). Answer: (a)

Teacher's Note:
1. Solve with the integrating factor \( \\sec x \).
2. Evaluate each option with \( y=x^2+\cos x \).

Answer: (b) \( y^2+2x^3+Cx^2=0 \)

Explanation:
1. Rewrite as \( xy\frac{dy}{dx}-y^2=-x^3 \). Put \( t=y^2 \): \( \frac x2\frac{dt}{dx}-t=-x^3 \), i.e. \( \frac{dt}{dx}-\frac2xt=-2x^2 \).
2. The integrating factor is \( \frac1{x^2} \): \( \frac t{x^2}=-2\int1\,dx=-2x+\lambda \).
3. So \( y^2+2x^3-\lambda x^2=0 \), i.e. \( y^2+2x^3+Cx^2=0 \). Answer: (b)

Teacher's Note:
1. Substitute \( t=y^2 \) to make the equation linear.
2. Use the integrating factor \( 1/x^2 \).

Answer: (c) \( \frac{49}{16} \)

Explanation:
1. \( \frac{dy}{dx}+\frac2xy=x \) has integrating factor \( x^2 \), so \( yx^2=\frac{x^4}4+C \). With \( y(1)=1 \), \( C=\frac34 \).
2. \( y=\frac{x^4+3}{4x^2} \), so \( y\left(\frac12\right)=\frac{\frac1{16}+3}{4\cdot\frac14}=\frac{49}{16} \). Answer: (c)

Teacher's Note:
1. Solve the linear equation.
2. Substitute \( x=1/2 \).

Answer: (a) \( \frac13+e^6 \)

Explanation:
1. The integrating factor is \( e^{3\tan x} \), so \( ye^{3\tan x}=\int e^{3\tan x}\sec^2x\,dx=\frac13e^{3\tan x}+C \), i.e. \( y=\frac13+Ce^{-3\tan x} \).
2. \( y\left(\frac\pi4\right)=\frac43\Rightarrow\frac13+Ce^{-3}=\frac43\Rightarrow C=e^3 \).
3. \( y\left(-\frac\pi4\right)=\frac13+e^3\cdot e^{3}=\frac13+e^6 \). Answer: (a)

Teacher's Note:
1. Find the integrating factor \( e^{3\\tan x} \).
2. Determine \( C \) from the given value.

Answer: (b) a circle with centre on the X-axis

Explanation:
1. The equation is homogeneous. Put \( y=vx \): \( v+x\frac{dv}{dx}=\frac{v^2-1}{2v} \), so \( x\frac{dv}{dx}=-\frac{1+v^2}{2v} \).
2. Separating: \( \int\frac{2v\,dv}{1+v^2}=-\int\frac{dx}x \), so \( (1+v^2)Cx=1 \), i.e. \( C(x^2+y^2)=x \).
3. Through \( (1,1) \): \( C=\frac12 \), giving \( x^2+y^2-2x=0 \), a circle with centre \( (1,0) \) on the X-axis. Answer: (b)

Teacher's Note:
1. Substitute \( y=vx \).
2. Separate variables and integrate.

Answer: (d) \( -\log_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \)

Explanation:
1. Put \( t=x-y \): \( 1-\frac{dt}{dx}=t^2 \), so \( \frac{dt}{dx}=1-t^2 \) and \( \int\frac{dt}{1-t^2}=\int dx \).
2. \( \frac12\log_e\left|\frac{1+t}{1-t}\right|=x+C \), i.e. \( \frac12\log_e\left|\frac{1+x-y}{1-x+y}\right|=x+C \).
3. \( y(1)=1\Rightarrow t=0\Rightarrow0=1+C\Rightarrow C=-1 \). So \( -\log_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \). Answer: (d)

Teacher's Note:
1. Substitute \( t=x-y \).
2. Use \( \int\frac{dt}{1-t^2}=\frac12\log\frac{1+t}{1-t} \).

Answer: (a) \( (\sqrt3,0) \)

Explanation:
1. \( \frac{dy}{dx}+\frac2xy=x \) has integrating factor \( x^2 \), so \( yx^2=\frac{x^4}4+C \).
2. Through \( (1,-2) \): \( -2=\frac14+C\Rightarrow C=-\frac94 \), so the curve is \( 4yx^2=x^4-9 \).
3. Testing the options, only \( (\sqrt3,0) \) satisfies it: \( 0=9-9 \). Answer: (a)

Teacher's Note:
1. Solve the linear equation.
2. Test each option in the equation of the curve.

Answer: (a) \( \frac13 \)

Explanation:
1. The equation says \( \frac{d}{dx}\big[(y+1)(2+\sin x)\big]=0 \), so \( (y+1)(2+\sin x)=k \).
2. At \( x=0 \): \( (2)(2)=4 \), so \( k=4 \) and \( y=\frac4{2+\sin x}-1 \).
3. At \( x=\frac\pi2 \): \( y=\frac43-1=\frac13 \). Answer: (a)

Teacher's Note:
1. Recognise a product derivative: \( (y+1)(2+\\sin x) \).
2. Use \( y(0)=1 \) to find the constant.

Answer: (d) \( \frac45 \)

Explanation:
1. \( y\,dx+xy^2dx=x\,dy\Rightarrow\frac{x\,dy-y\,dx}{y^2}=x\,dx\Rightarrow-d\left(\frac xy\right)=x\,dx \).
2. Integrating: \( -\frac xy=\frac{x^2}2+C \). Through \( (1,-1) \): \( 1=\frac12+C\Rightarrow C=\frac12 \).
3. So \( y=-\frac{2x}{x^2+1} \) and \( f\left(-\frac12\right)=\frac{1}{5/4}=\frac45 \). Answer: (d)

Teacher's Note:
1. Rearrange to get \( d(x/y) \).
2. Integrate and use the point.

Answer: (a) \( 400-300e^{t/2} \)

Explanation:
1. \( \frac{dp}{dt}-\frac12p=-200 \) has integrating factor \( e^{-t/2} \), so \( p\,e^{-t/2}=-200\int e^{-t/2}dt=400e^{-t/2}+K \).
2. \( p(0)=100\Rightarrow100=400+K\Rightarrow K=-300 \), so \( p(t)=400-300e^{t/2} \). Answer: (a)

Teacher's Note:
1. Use an integrating factor.
2. Use \( p(0)=100 \).

Answer: (c) 3500

Explanation:
1. \( P=\int(100-12\sqrt x)dx=100x-8x^{3/2}+C \). At \( x=0 \), \( P=2000 \), so \( C=2000 \).
2. At \( x=25 \): \( P=2500-8\cdot125+2000=3500 \). Answer: (c)

Teacher's Note:
1. Integrate the rate.
2. Use \( P(0)=2000 \).

Answer: (d) 7

Explanation:
1. \( \int\frac{dy}{y+3}=\int dx\Rightarrow\log|y+3|=x+C \). At \( x=0 \): \( \log5=C \).
2. At \( x=\log2 \): \( \log(y+3)=\log2+\log5=\log10 \), so \( y+3=10 \) and \( y=7 \). Answer: (d)

Teacher's Note:
1. Separate the variables.
2. Use \( y(0)=2 \).

Answer: (a) \( I-\frac k2T^2 \)

Explanation:
1. \( dV=-k(T-t)\,dt \), with \( V(0)=I \).
2. \( V(T)-I=-k\int_0^T(T-t)\,dt=-k\cdot\frac{T^2}2 \), so \( V(T)=I-\frac k2T^2 \). Answer: (a)

Teacher's Note:
1. Integrate from \( t=0 \) to \( t=T \).
2. Use \( V(0)=I \).

Answer: (a) \( \sec x=(\tan x+C)y \)

Explanation:
1. \( \frac{dy}{dx}=y\tan x-y^2\sec x \) is a Bernoulli equation. Divide by \( y^2 \): \( \frac1{y^2}\frac{dy}{dx}-\frac1y\tan x=-\sec x \).
2. Put \( z=\frac1y \): \( \frac{dz}{dx}+(\tan x)z=\sec x \), with integrating factor \( \sec x \). So \( z\sec x=\int\sec^2x\,dx=\tan x+C \).
3. \( \frac{\sec x}y=\tan x+C \), i.e. \( \sec x=(\tan x+C)y \). Answer: (a)

Teacher's Note:
1. Recognise a Bernoulli equation with \( n=2 \).
2. Substitute \( z=1/y \).

Answer: (d) \( yy''=(y')^2 \)

Explanation:
1. \( y'=c_2c_1e^{c_2x}=c_2y \), so \( c_2=\frac{y'}y \).
2. \( y''=c_2y'=\frac{(y')^2}y \), i.e. \( yy''=(y')^2 \). Answer: (d)

Teacher's Note:
1. Differentiate twice.
2. Eliminate \( c_1,c_2 \).

Answer: (a) \( y=x\log x+x \)

Explanation:
1. \( \frac{dy}{dx}-\frac1xy=1 \) has integrating factor \( e^{-\log x}=\frac1x \).
2. \( \frac yx=\int\frac1x\,dx=\log x+C \). \( y(1)=1\Rightarrow C=1 \), so \( y=x\log x+x \). Answer: (a)

Teacher's Note:
1. Write in linear form.
2. Use the integrating factor \( 1/x \).

Answer: (c) \( y^2=x^2+2xy\frac{dy}{dx} \)

Explanation:
1. The family is \( x^2+y^2+2gx=0 \). Differentiate: \( 2x+2yy'+2g=0 \), so \( 2g=-(2x+2yy') \).
2. Substitute: \( x^2+y^2-(2x+2yy')x=0 \), i.e. \( y^2=x^2+2xyy' \). Answer: (c)

Teacher's Note:
1. Write the general circle through the origin with centre on the X-axis.
2. Eliminate \( g \).

Answer: (b) order 1, degree 3

Explanation:
1. Differentiate: \( 2yy'=2c\Rightarrow c=yy' \).
2. Substitute: \( y^2=2yy'(x+\sqrt{yy'}) \), so \( (y^2-2xyy')^2=4(yy')^3 \) after removing the radical.
3. The highest derivative \( y' \) appears to the power 3, so the order is 1 and the degree is 3. Answer: (b)

Teacher's Note:
1. Find \( c \) in terms of \( y,y' \).
2. Remove the radical before reading the degree.

Answer: (b) \( \log\left(\frac yx\right)=Cx \)

Explanation:
1. \( \frac{dy}{dx}=\frac yx\left(\log\frac yx+1\right) \) is homogeneous. Put \( y=tx \): \( t+x\frac{dt}{dx}=t\log t+t \).
2. \( x\frac{dt}{dx}=t\log t \), so \( \int\frac{dt}{t\log t}=\int\frac{dx}x \) and \( \log(\log t)=\log x+\log C \).
3. \( \log t=Cx \), i.e. \( \log\frac yx=Cx \). Answer: (b)

Teacher's Note:
1. Substitute \( y=tx \).
2. Separate the variables.

Answer: (c) \( (x^2-y^2)y'=2xy \)

Explanation:
1. Differentiate: \( 2x+2yy'-2ay'=0 \), so \( 2a=\frac{2x+2yy'}{y'} \). From the family, \( 2a=\frac{x^2+y^2}y \).
2. Equate: \( y(2x+2yy')=(x^2+y^2)y' \), so \( 2xy=(x^2-y^2)y' \). Answer: (c)

Teacher's Note:
1. Differentiate and find \( 2a \).
2. Equate with the value from the original equation.

Answer: (b) \( -\frac1{xy}+\log y=C \)

Explanation:
1. Divide by \( x^2y^2 \): \( \frac{y\,dx+x\,dy}{x^2y^2}=-\frac1y\,dy \), and the left side is \( d\left(-\frac1{xy}\right) \).
2. Integrate: \( -\frac1{xy}=-\log y+C \), i.e. \( -\frac1{xy}+\log y=C \). Answer: (b)

Teacher's Note:
1. Group \( y\\,dx+x\\,dy=d(xy) \).
2. Divide by \( (xy)^2 \).

Answer: (b) 1, 2

Explanation:
1. The family is \( y^2=4a(x+h) \). Differentiate: \( 2yy'=4a \), i.e. \( yy'=2a \).
2. Differentiate again: \( (y')^2+yy''=0 \). Its highest derivative \( y'' \) has power 1 and order 2. So degree 1 and order 2. Answer: (b)

Teacher's Note:
1. Two parameters need two differentiations.
2. Read off the power of the highest derivative.

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

Explanation:
1. Write as \( \frac{dx}{dy}+\frac x{1+y^2}=\frac{e^{\tan^{-1}y}}{1+y^2} \), linear in \( x \) with integrating factor \( e^{\tan^{-1}y} \).
2. \( xe^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^2}dy=\frac12e^{2\tan^{-1}y}+C_1 \).
3. So \( 2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+C \). Answer: (b)

Teacher's Note:
1. Treat \( x \) as a function of \( y \).
2. Substitute \( t=\tan^{-1}y \).

Answer: (c) (3, 3)

Explanation:
1. Cube both sides to remove the fractional power: \( \left(1+3\frac{dy}{dx}\right)^2=64\left(\frac{d^3y}{dx^3}\right)^3 \).
2. The highest order derivative is \( \frac{d^3y}{dx^3} \), so the order is 3, and its power is 3, so the degree is 3. Answer: (c)

Teacher's Note:
1. Remove the fractional power first.
2. Order is the highest derivative; degree is its power.

Answer: (b) \( \frac{e^{-2x}}4+Cx+D \)

Explanation:
1. Integrate once: \( \frac{dy}{dx}=-\frac{e^{-2x}}2+C \).
2. Integrate again: \( y=\frac{e^{-2x}}4+Cx+D \). Answer: (b)

Teacher's Note:
1. Integrate twice.
2. Include one constant per integration.

Answer: (a) \( \frac{d^2y}{dx^2}=0 \)

Explanation:
1. A non-vertical line has the form \( ax+by=1 \) with \( b\ne0 \). Differentiate: \( a+b\frac{dy}{dx}=0 \).
2. Differentiate again: \( b\frac{d^2y}{dx^2}=0 \), so \( \frac{d^2y}{dx^2}=0 \) since \( b\ne0 \). Answer: (a)

Teacher's Note:
1. Write the general non-vertical line.
2. Differentiate twice.