JEE Mathematics Differential Equations MCQs Set 11

Answer: (a) \( \frac{dr}{d\theta}=-r\tan2\theta \)

Explanation:
1. Differentiate: \( 2r\frac{dr}{d\theta}=-2a^2\sin2\theta \).
2. Substitute \( a^2=\frac{r^2}{\cos2\theta} \): \( \frac{dr}{d\theta}=-r\tan2\theta \).

Teacher's Note:
1. Write the family with its parameters, differentiate repeatedly and eliminate the parameters.
2. Check the final answer by differentiating the given solution once more.

Answer: (c) \( (y-2)^2(y')^2=25-(y-2)^2 \)

Explanation:
1. The circles are \( (x-a)^2+(y-2)^2=25 \).
2. Differentiating, \( (x-a)+(y-2)y'=0 \), so \( (x-a)^2=(y-2)^2y'^2 \).
3. Substituting back: \( (y-2)^2y'^2+(y-2)^2=25 \).
4. Rearranged: \( (y-2)^2y'^2=25-(y-2)^2 \).

Teacher's Note:
1. Differentiate as many times as there are arbitrary constants, then eliminate them.
2. The order of the resulting equation equals the number of independent constants.

Answer: (b) \( y^{2/3}-x^{2/3}=c \)

Explanation:
1. \( y^{-1/3}dy=x^{-1/3}dx \).
2. \( \frac32y^{2/3}=\frac32x^{2/3}+k \).
3. So \( y^{2/3}-x^{2/3}=c \).

Teacher's Note:
1. Look for a factorisation of the right side as f(x) times g(y).
2. Standard integrals such as dx/(1+x²) appear again and again.

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

Explanation:
1. \( \frac{dy}{\sqrt{1+y^2}}+\frac{dx}{\sqrt{1+x^2}}=0 \).
2. \( \int\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}x \).
3. So \( \sinh^{-1}x+\sinh^{-1}y=c \).

Teacher's Note:
1. Separate the variables so that each side has only one variable, then integrate both sides.
2. Add the constant of integration once and simplify it into a convenient form.

Answer: (c) \( y=\log|e^x+e^{-x}|+c \)

Explanation:
1. \( dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx \).
2. The numerator is the derivative of the denominator.
3. \( y=\log(e^x+e^{-x})+c \).

Teacher's Note:
1. Look for a factorisation of the right side as f(x) times g(y).
2. Standard integrals such as dx/(1+x²) appear again and again.

Answer: (a) \( y=c(1+x^2) \)

Explanation:
1. \( \frac{dy}{y}=\frac{2x}{1+x^2}dx \).
2. \( \log y=\log(1+x^2)+\log c \).
3. So \( y=c(1+x^2) \).

Teacher's Note:
1. Separate the variables so that each side has only one variable, then integrate both sides.
2. Add the constant of integration once and simplify it into a convenient form.

Answer: (a) \( e^x-e^y=c \)

Explanation:
1. \( e^y\,dy=e^x\,dx \).
2. \( e^y=e^x+k \).
3. So \( e^x-e^y=c \).

Teacher's Note:
1. Look for a factorisation of the right side as f(x) times g(y).
2. Standard integrals such as dx/(1+x²) appear again and again.

Answer: (a) \( \sec y=c(1+x^2) \)

Explanation:
1. \( \tan y\,dy=\frac{2x}{1+x^2}dx \).
2. \( -\log\cos y=\log(1+x^2)+\log c \).
3. \( \sec y=c(1+x^2) \).

Teacher's Note:
1. Separate the variables so that each side has only one variable, then integrate both sides.
2. Add the constant of integration once and simplify it into a convenient form.

Answer: (a) \( y-x=c(1+xy) \)

Explanation:
1. \( \tan^{-1}y=\tan^{-1}x+k \).
2. Taking tan of both sides, \( \frac{y-x}{1+xy}=c \).
3. So \( y-x=c(1+xy) \).

Teacher's Note:
1. Look for a factorisation of the right side as f(x) times g(y).
2. Standard integrals such as dx/(1+x²) appear again and again.

Answer: (a) \( y=3x \)

Explanation:
1. \( \frac{dy}{y}=\frac{dx}{x} \), so \( y=cx \).
2. At (1, 3): c = 3.
3. So \( y=3x \).

Teacher's Note:
1. Separate the variables so that each side has only one variable, then integrate both sides.
2. Add the constant of integration once and simplify it into a convenient form.

Answer: (b) \( e^y=e^x+\frac{x^3}{3}+c \)

Explanation:
1. Since \( e^{2\log x}=x^2 \), \( \frac{dy}{dx}=e^{-y}(e^x+x^2) \).
2. \( e^y\,dy=(e^x+x^2)dx \).
3. \( e^y=e^x+\frac{x^3}{3}+c \).

Teacher's Note:
1. Look for a factorisation of the right side as f(x) times g(y).
2. Standard integrals such as dx/(1+x²) appear again and again.

Answer: (c) \( \sin x\sin y=c \)

Explanation:
1. \( \frac{dy}{\tan y}+\frac{dx}{\tan x}=0 \).
2. \( \log\sin y+\log\sin x=\log c \).
3. So \( \sin x\sin y=c \).

Teacher's Note:
1. Separate the variables so that each side has only one variable, then integrate both sides.
2. Add the constant of integration once and simplify it into a convenient form.

Answer: (a) \( x\sqrt{1+x^2}+y\sqrt{1+y^2}+\log\left[\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)\right]=c \)

Explanation:
1. Use \( \int\sqrt{1+x^2}dx=\frac12\left[x\sqrt{1+x^2}+\log\left(x+\sqrt{1+x^2}\right)\right] \).
2. Apply the same formula in y and add.
3. Multiply by 2 and combine the logarithms.

Teacher's Note:
1. Look for a factorisation of the right side as f(x) times g(y).
2. Standard integrals such as dx/(1+x²) appear again and again.

Answer: (b) \( \sqrt{1+x^2}+\sqrt{1+y^2}=c \)

Explanation:
1. Divide by \( \sqrt{(1+x^2)(1+y^2)} \): \( \frac{x\,dx}{\sqrt{1+x^2}}+\frac{y\,dy}{\sqrt{1+y^2}}=0 \).
2. Integrating, \( \sqrt{1+x^2}+\sqrt{1+y^2}=c \).

Teacher's Note:
1. Separate the variables so that each side has only one variable, then integrate both sides.
2. Add the constant of integration once and simplify it into a convenient form.

Answer: (b) \( y\frac{dv}{dy}+\frac{v+e^v}{1+e^v}=0 \)

Explanation:
1. Put \( x=vy,\ dx=v\,dy+y\,dv \): \( (1+e^v)(v\,dy+y\,dv)+e^v(1-v)dy=0 \).
2. \( (v+e^v)dy+y(1+e^v)dv=0 \).
3. So \( y\frac{dv}{dy}+\frac{v+e^v}{1+e^v}=0 \).

Teacher's Note:
1. A homogeneous equation has every term of the same degree in x and y.
2. The substitution y = vx (or x = vy) always reduces it to a separable form.

Answer: (b) \( \sin^{-1}\left(\frac yx\right)=\log(cx) \)

Explanation:
1. Put \( y=vx \): \( v+x\frac{dv}{dx}=\sqrt{1-v^2}+v \).
2. \( \frac{dv}{\sqrt{1-v^2}}=\frac{dx}{x} \).
3. \( \sin^{-1}v=\log x+\log c \), so \( \sin^{-1}\frac yx=\log(cx) \).

Teacher's Note:
1. Put y = vx so that dy/dx = v + x dv/dx and the variables separate.
2. Return to y/x after integrating.

Answer: (b) \( x^2-y^2-2xy+4x+2y=c \)

Explanation:
1. \( (x-y+2)dx-(x+y-1)dy=0 \) has \( M_y=-1=N_x \), so it is exact.
2. \( F=\frac{x^2}{2}-xy+2x+g(y) \), \( F_y=-x+g'=-(x+y-1) \) gives \( g=-\frac{y^2}{2}+y \).
3. So \( x^2-y^2-2xy+4x+2y=c \).

Teacher's Note:
1. For (ax+by+c)/(a'x+b'y+c') find the intersection of the lines first.
2. If the lines are parallel use v = ax + by.

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

Explanation:
1. Put \( u=x-2y \): \( \frac{du}{dx}=1-2\cdot\frac{u+1}{2u}=-\frac1u \).
2. \( u\,du=-dx \).
3. \( \frac{u^2}{2}=-x+k \), so \( (x-2y)^2+2x=c \).

Teacher's Note:
1. Shift the origin to the point of intersection of the two lines, or use exactness if it applies.
2. After shifting, the equation becomes homogeneous.

Answer: (a) \( 1+x^3 \)

Explanation:
1. Divide by \( 1+x^3 \): \( P=\frac{3x^2}{1+x^3} \).
2. \( \int P\,dx=\log(1+x^3) \).
3. IF \( =1+x^3 \).

Teacher's Note:
1. Bring the equation to standard linear form before reading P.
2. If the equation is linear in x, use dx/dy + Px = Q and e raised to the integral of P dy.

Answer: (b) \( \frac{e^x}{x} \)

Explanation:
1. Rewrite: \( \frac{dy}{dx}+y\left(1-\frac1x\right)=\frac1x \).
2. \( \int\left(1-\frac1x\right)dx=x-\log x \).
3. IF \( =\frac{e^x}{x} \).

Teacher's Note:
1. Write the equation as dy/dx + Py = Q; the integrating factor is e raised to the integral of P dx.
2. Constants in the integral of P are not needed.

Answer: (b) \( e^{\tan x} \)

Explanation:
1. Divide by \( \cos^2x \): \( P=\sec^2x \).
2. \( \int\sec^2x\,dx=\tan x \).
3. IF \( =e^{\tan x} \).

Teacher's Note:
1. Bring the equation to standard linear form before reading P.
2. If the equation is linear in x, use dx/dy + Px = Q and e raised to the integral of P dy.

Answer: (b) \( \cot x \)

Explanation:
1. IF \( =e^{\int P\,dx}=\sin x \).
2. So \( \int P\,dx=\log\sin x \).
3. Differentiating, \( P=\cot x \).

Teacher's Note:
1. Bring the equation to standard linear form before reading P.
2. If the equation is linear in x, use dx/dy + Px = Q and e raised to the integral of P dy.

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

Explanation:
1. IF \( =e^{2\log x}=x^2 \).
2. \( x^2y=\int5x^2\cdot x^2dx=\int5x^4dx=x^5+c \).

Teacher's Note:
1. Multiply the linear equation by its integrating factor so the left side becomes an exact derivative.
2. Then integrate and divide by the factor.