JEE Mathematics Differential Equations MCQs Set 19

Answer: (d) \( y^2=1-\dfrac{x}{\sec x+\tan x} \)

Explanation:
1. Multiply by \(2y\): \( 2y\frac{dy}{dx}+y^2\sec x=\tan x \). Put \( u=y^2 \): \( \frac{du}{dx}+u\sec x=\tan x \), a linear equation.
2. The integrating factor is \( e^{\int\sec x\,dx}=\sec x+\tan x \). So \( u(\sec x+\tan x)=\int\tan x(\sec x+\tan x)dx=\sec x+\tan x-x+C \).
3. \( y(0)=1 \) gives \( C=0 \), so \( y^2=1-\frac{x}{\sec x+\tan x} \).

Teacher's Note:
1. Substituting u = y^2 turns the equation into a first-order linear one.

Answer: (d) 0

Explanation:
1. \( \frac{dy}{dx}=\frac{x^2+4x-9}{x+2}=x+2-\frac{13}{x+2} \).
2. Integrating: \( y=\frac{x^2}2+2x-13\ln|x+2|+C \). \( y(0)=0 \) gives \( C=13\ln2 \).
3. \( y(-4)=8-8-13\ln2+13\ln2=0 \).

Teacher's Note:
1. Divide the polynomial by (x + 2) before integrating.

Answer: (b) \( 2x \)

Explanation:
1. The circles are \( x^2+y^2+2ky=0 \). Differentiating: \( 2x+2yy'+2ky'=0 \), so \( k=-\frac{x+yy'}{y'} \).
2. Substituting: \( x^2+y^2-\frac{2y(x+yy')}{y'}=0 \), i.e. \( \left(x^2+y^2\right)y'-2xy-2y^2y'=0 \).
3. So \( \left(x^2-y^2\right)y'=2xy \) and \( g(x)=2x \).

Teacher's Note:
1. A circle touching the x-axis at the origin has its centre on the y-axis and passes through (0, 0).

Answer: (d) \( y\sqrt{\cot x}=x+c \)

Explanation:
1. Write \( \frac{dy}{dx}-\frac{y}{\sin2x}=\sqrt{\tan x} \). The integrating factor is \( e^{-\int\operatorname{cosec}2x\,dx}=e^{-\frac12\ln\tan x}=\frac1{\sqrt{\tan x}} \).
2. \( \frac{d}{dx}\left(\frac{y}{\sqrt{\tan x}}\right)=1 \), so \( \frac y{\sqrt{\tan x}}=x+c \).
3. That is \( y\sqrt{\cot x}=x+c \).

Teacher's Note:
1. Divide by sin 2x to put the equation in the linear form, then use the integrating factor.

Answer: (c) \( -5 \)

Explanation:
1. The integrating factor is \( e^{\int\tan x\,dx}=\sec x \). \( \frac{d}{dx}(y\sec x)=\sin2x\sec x=2\sin x \).
2. \( y\sec x=-2\cos x+C \); \( y(0)=1 \) gives \( C=3 \), so \( y=3\cos x-2\cos^2x \).
3. \( y(\pi)=-3-2=-5 \).

Teacher's Note:
1. The integrating factor sec x simplifies sin 2x sec x to 2 sin x.

Answer: (c) 3, degree is not defined

Explanation:
1. The highest derivative is \( \frac{d^3y}{dx^3} \), so the order is 3.
2. The equation is not a polynomial in the derivatives because the third derivative appears in an exponential.
3. Hence the degree is not defined.

Teacher's Note:
1. Degree exists only when the equation is a polynomial in the derivatives.

Answer: (c) \( xy_2+2y_1=xy \)

Explanation:
1. Let \( u=xy \). Then \( u=ae^{-x}+be^x \) gives \( u''=u \).
2. \( u''=(xy)''=xy''+2y' \).
3. So \( xy_2+2y_1=xy \).

Teacher's Note:
1. Treat xy as a single function u with u'' = u.

Answer: (a) 2, 2

Explanation:
1. Multiply by \( \frac{d^2y}{dx^2} \): \( r\frac{d^2y}{dx^2}=\left[r+\left(\frac{dy}{dx}\right)^2\right]^{3/2} \).
2. Squaring to clear the fractional power: \( r^2\left(\frac{d^2y}{dx^2}\right)^2=\left[r+\left(\frac{dy}{dx}\right)^2\right]^3 \).
3. The highest derivative is of order 2 and is raised to the power 2, so the order is 2 and the degree is 2.

Teacher's Note:
1. Clear the fractional power first, then read off the power of the highest derivative.

Answer: (a) Three

Explanation:
1. Differentiating, \( \frac{dy}{dx}=A \).
2. Eliminating \(A\): \( y=x\frac{dy}{dx}+\left(\frac{dy}{dx}\right)^3 \).
3. The highest derivative \( \frac{dy}{dx} \) has power 3, so the degree is 3.

Teacher's Note:
1. Eliminate the constant using y' = A.

Answer: (b) \( \left[1+\left(\frac{dy}{dx}\right)^2\right]^3=a^2\left(\frac{d^2y}{dx^2}\right)^2 \)

Explanation:
1. Differentiate twice: \( (x-h)+(y-k)y'=0 \) and \( 1+y'^2+(y-k)y''=0 \).
2. So \( y-k=-\frac{1+y'^2}{y''} \) and \( x-h=\frac{y'\left(1+y'^2\right)}{y''} \).
3. Substituting: \( \frac{\left(1+y'^2\right)^2\left(y'^2+1\right)}{y''^2}=a^2 \), i.e. \( \left[1+y'^2\right]^3=a^2y''^2 \).

Teacher's Note:
1. The radius of curvature formula gives the same relation directly.

Answer: (a) \( 2e^2 \)

Explanation:
1. \( \frac{\phi'}{\phi}=1 \), so \( \phi(x)=Ce^x \).
2. \( \phi(1)=2 \) gives \( C=\frac2e \).
3. \( \phi(3)=\frac2e\cdot e^3=2e^2 \).

Teacher's Note:
1. A function equal to its own derivative is a constant multiple of e^x.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. Assertion: the equation is \( dx-y\,dy+\sqrt{x^2+y^2}(x\,dx+y\,dy)=0 \). Integrating, \( x-\frac{y^2}2+\frac13\left(x^2+y^2\right)^{3/2}=\text{constant} \), so the Assertion is true.
2. Reason: \( y\,dx+x\,dy+xy(y\,dx-x\,dy)=0 \). Divide by \( x^2y^2 \): \( \frac{d(xy)}{(xy)^2}+\frac{dx}x-\frac{dy}y=0 \), giving \( -\frac1{xy}+\ln\frac xy=c \): the Reason is true.
3. The Reason concerns a different equation, so it does not explain the Assertion.

Teacher's Note:
1. Both statements are verified independently; the second equation has nothing to do with the first.

Answer: (a) \( y^2=kx \)

Explanation:
1. The tangent at \( P(x,y) \) meets the x-axis where \( X=x-\frac y{y'} \).
2. The y-axis point is the midpoint of P and this point, so \( \frac{x+\left(x-\frac y{y'}\right)}2=0 \), i.e. \( \frac y{y'}=2x \).
3. Then \( \frac{dy}y=\frac{dx}{2x} \) gives \( y^2=kx \).

Teacher's Note:
1. Use the midpoint condition on the x-coordinates to get y/y' = 2x.

Answer: (b) \( y_1(x)\displaystyle\int\frac{r(x)}{y_1(x)}dx \)

Explanation:
1. Try \( y=y_1u \). Then \( y'+fy=y_1'u+y_1u'+fy_1u=u\left(y_1'+fy_1\right)+y_1u'=y_1u' \).
2. For this to equal \( r \), \( u'=\frac r{y_1} \).
3. So a solution is \( y=y_1\int\frac{r}{y_1}dx \).

Teacher's Note:
1. This is variation of parameters applied to a first-order linear equation.

Answer: (b) 1

Explanation:
1. Separate: \( \frac{dy}{y(1-y)}=dx \), i.e. \( \left(\frac1y+\frac1{1-y}\right)dy=dx \).
2. \( \ln\left|\frac y{1-y}\right|=x+C \), so \( \frac{y}{y-1}=Ke^x \) with \( K>0 \).
3. \( y(0)=2 \) gives \( K=2 \), so \( y=\frac{2e^x}{2e^x-1}\to1 \) as \( x\to\infty \).

Teacher's Note:
1. y = 1 is a stable equilibrium of the logistic equation, and y(0) = 2 lies above it.

Answer: (c) \( \tan^{-1}y+\sin^{-1}x=c \)

Explanation:
1. Separate the variables: \( \frac{dy}{1+y^2}=-\frac{dx}{\sqrt{1-x^2}} \).
2. Integrating: \( \tan^{-1}y=-\sin^{-1}x+c \).
3. So \( \tan^{-1}y+\sin^{-1}x=c \).

Teacher's Note:
1. The variables separate directly.

Answer: (d) \( y=-f(x)+ce^xf(x) \)

Explanation:
1. Divide by \( f \): \( y'-\left(1+\frac{f'}f\right)y=f \). The integrating factor is \( e^{-x}\cdot\frac1f \).
2. \( \frac{d}{dx}\left(\frac{ye^{-x}}{f}\right)=e^{-x} \), so \( \frac{ye^{-x}}f=-e^{-x}+c \).
3. \( y=-f(x)+cf(x)e^x \).

Teacher's Note:
1. The integrating factor is exp(-x - ln f) = e^(-x)/f.

Answer: (a) \( \frac1{3x}+\frac{2x^2}3 \)

Explanation:
1. The limit has the form \( \frac00 \); by L'Hopital's rule (differentiate with respect to \(t\)): \( 2xf(x)-x^2f'(x)=1 \).
2. So \( f'-\frac2xf=-\frac1{x^2} \). With integrating factor \( x^{-2} \): \( \frac{d}{dx}\left(\frac f{x^2}\right)=-x^{-4} \), so \( \frac f{x^2}=\frac1{3x^3}+C \).
3. \( f=\frac1{3x}+Cx^2 \); \( f(1)=1 \) gives \( C=\frac23 \).

Teacher's Note:
1. The limit is the derivative of t^2 f(x) - x^2 f(t) with respect to t, evaluated at t = x.

Answer: (a) \( \sin\left(\frac yx\right)=\log x+\frac12 \)

Explanation:
1. Put \( y=vx \): \( v+x\frac{dv}{dx}=v+\sec v \), so \( \cos v\,dv=\frac{dx}x \).
2. \( \sin v=\ln x+c \). At \( (1,\frac\pi6) \): \( v=\frac\pi6 \), \( \sin v=\frac12 \), so \( c=\frac12 \).
3. The curve is \( \sin\frac yx=\log x+\frac12 \).

Teacher's Note:
1. A homogeneous equation: substitute y = vx.

Answer: (d) Order 1, degree 3

Explanation:
1. There is one parameter, so the order is 1. Differentiating, \( 2yy'=2c \), so \( c=yy' \).
2. Substituting: \( y^2=2yy'\left(x+\sqrt{yy'}\right) \), so \( y^2-2xyy'=2yy'\sqrt{yy'} \).
3. Squaring: \( \left(y^2-2xyy'\right)^2=4y^3y'^3 \). The highest power of \( y' \) is 3, so the degree is 3.

Teacher's Note:
1. Remove the radical by squaring after isolating it.

Answer: (a) \( I-\frac{kT^2}2 \)

Explanation:
1. Integrate: \( V(t)=-k\left(Tt-\frac{t^2}2\right)+C \). \( V(0)=I \) gives \( C=I \).
2. \( V(T)=I-k\left(T^2-\frac{T^2}2\right) \).
3. \( V(T)=I-\frac{kT^2}2 \).

Teacher's Note:
1. The scrap value is the value at t = T, the end of the life.

Answer: (d) \( 5-\ln2 \)

Explanation:
1. \( \frac{dy}{dx}-y=x \). The integrating factor is \( e^{-x} \): \( \frac{d}{dx}\left(ye^{-x}\right)=xe^{-x} \).
2. \( ye^{-x}=-xe^{-x}-e^{-x}+C \), so \( y=-x-1+Ce^x \). \( y(0)=2 \) gives \( C=3 \).
3. \( y(\ln2)=-\ln2-1+6=5-\ln2 \).

Teacher's Note:
1. Solve the linear equation with integrating factor e^(-x) and use the initial value.

Answer: (d) \( y=\frac6x \)

Explanation:
1. The tangent at \( P(x,y) \) meets the axes at \( \left(x-\frac y{y'},0\right) \) and \( (0,y-xy') \). P is the midpoint, so \( x=\frac12\left(x-\frac y{y'}\right) \), giving \( x\,y'=-y \).
2. Then \( \frac{dy}y=-\frac{dx}x \), so \( xy=c \).
3. Through \( (2,3) \): \( c=6 \), so \( y=\frac6x \).

Teacher's Note:
1. Midpoint of the intercepts gives x y' + y = 0, whose solution is a rectangular hyperbola.

Answer: (d) 3

Explanation:
1. The general solution has three independent arbitrary constants \( a,b,c \).
2. The order equals the number of independent arbitrary constants.
3. So the order is 3.

Teacher's Note:
1. Count the independent arbitrary constants.

Answer: (a) 10

Explanation:
1. Shift the origin to the intersection of \( 3y-7x-3=0 \) and \( 3x-7y+7=0 \), which is \( (0,1) \): put \( Y=y-1 \). Then \( \frac{dY}{dx}=\frac{3Y-7x}{3x-7Y} \), a homogeneous equation.
2. With \( Y=vx \): \( \frac{(3-7v)\,dv}{7(v^2-1)}=\frac{dx}x \), and \( \frac{3-7v}{(v-1)(v+1)}=\frac{-2}{v-1}+\frac{-5}{v+1} \).
3. Integrating: \( (v-1)^2(v+1)^5x^7=\text{constant} \), i.e. \( (Y-x)^2(Y+x)^5=\text{constant} \). So \( m=2,\ n=5 \) and \( mn=10 \).

Teacher's Note:
1. Shift the origin to the point where both numerator and denominator vanish, then substitute Y = vx.

Answer: (b) 4

Explanation:
1. Multiply by \( \frac{d^3y}{dx^3} \): \( \left(\frac{d^2y}{dx^2}\right)^5\frac{d^3y}{dx^3}+4\left(\frac{d^2y}{dx^2}\right)^3=\left(x^2-1\right)\frac{d^3y}{dx^3} \).
2. The highest derivative is of order 3, so \( m=3 \), and it appears to the power 1, so \( n=1 \).
3. \( m+n=4 \).

Teacher's Note:
1. Clear the denominator before reading the degree.

Answer: (c) 1

Explanation:
1. Such parabolas are \( y^2=4a(x-h) \) with two constants.
2. Differentiating twice: \( 2yy'=4a \) and \( yy''+y'^2=0 \).
3. The equation \( yy''+y'^2=0 \) has the highest derivative \( y'' \) to the power 1, so the degree is 1.

Teacher's Note:
1. Two arbitrary constants mean order 2; here the constants vanish after two differentiations.

Answer: (a) 1

Explanation:
1. \( e^{-y}dy=\left(e^x+e^{-x}\right)dx \), so \( -e^{-y}=e^x-e^{-x}+C \).
2. Through \( (1,1) \): \( -e^{-1}=e-e^{-1}+C \), so \( C=-e \).
3. At \( x=0 \): \( -e^{-k}=0-e \), so \( e^{-k}=e \) and \( k=-1 \), \( |k|=1 \).

Teacher's Note:
1. Multiply by e^(-y) to separate the variables.

Answer: (b) 5

Explanation:
1. \(h,k\) satisfy \( h-2k+5=0 \) and \( 2h-k+4=0 \).
2. From the first, \( h=2k-5 \); the second gives \( 4k-10-k+4=0 \), so \( k=2,\ h=-1 \).
3. \( h^2+k^2=1+4=5 \).

Teacher's Note:
1. Solve the two linear equations formed by the numerator and denominator.