JEE Mathematics Differential Equations MCQs Set 13

Answer: (c) \( y=\sin\left(\frac cx\right) \)

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

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) \( x^2+3xy-2y^2-5x+y=c \)

Explanation:
1. Here \( M_y=3=N_x \), so the equation is exact.
2. \( F=x^2+3xy-5x+g(y) \) and \( F_y=3x+g'=3x-4y+1 \).
3. \( g=-2y^2+y \).
4. \( x^2+3xy-2y^2-5x+y=c \).

Teacher's Note:
1. If the equation is exact, the solution is F(x, y) = c with F_x = M and F_y = N.

Answer: (b) \( x^2-4xy+y^2+6x-10y=c \)

Explanation:
1. \( (x-2y+3)dx-(2x-y+5)dy=0 \) has \( M_y=-2=N_x \), so it is exact.
2. \( F=\frac{x^2}{2}-2xy+3x+g(y) \), \( F_y=-2x+g'=-(2x-y+5) \) gives \( g=\frac{y^2}{2}-5y \).
3. Multiply by 2: \( x^2-4xy+y^2+6x-10y=c \).

Teacher's Note:
1. Test exactness: partial of M with respect to y equals partial of N with respect to x.
2. Then integrate M with respect to x and add a function of y.

Answer: (a) \( 3x^2+7xy-2y^2-4x+3y=c \)

Explanation:
1. Here \( M_y=7=N_x \): exact.
2. \( F=3x^2+7xy-4x+g(y) \) and \( F_y=7x+g'=7x-4y+3 \).
3. \( g=-2y^2+3y \).
4. \( 3x^2+7xy-2y^2-4x+3y=c \).

Teacher's Note:
1. If the equation is exact, the solution is F(x, y) = c with F_x = M and F_y = N.

Answer: (a) \( xe^x \)

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

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: (c) \( x\sin x \)

Explanation:
1. Divide by \( x\sin x \): \( P=\cot x+\frac1x \).
2. \( \int P\,dx=\log\sin x+\log x \).
3. IF \( =x\sin x \).

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) \( x=2y^2+cy \)

Explanation:
1. Rewrite: \( \frac{dx}{dy}-\frac xy=2y \).
2. IF \( =\frac1y \).
3. \( \frac xy=\int2\,dy=2y+c \).
4. So \( x=2y^2+cy \).

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.

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

Explanation:
1. \( \frac{dy}{dx}-y=x^2 \).
2. IF \( =e^{-x} \).
3. \( ye^{-x}=\int x^2e^{-x}dx=-(x^2+2x+2)e^{-x}+c \).
4. So \( y+x^2+2x+2=ce^x \).

Teacher's Note:
1. Standard form dy/dx + Py = Q gives y times IF = integral of Q times IF.
2. Do not forget the constant of integration.

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

Explanation:
1. \( \frac{dy}{dx}+\left(\frac1x+\cot x\right)y=1 \).
2. IF \( =x\sin x \).
3. \( y\,x\sin x=\int x\sin x\,dx=-x\cos x+\sin x+c \).

Teacher's Note:
1. Standard form dy/dx + Py = Q gives y times IF = integral of Q times IF.
2. Do not forget the constant of integration.

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

Explanation:
1. \( \frac{dy}{dx}-\frac{y}{\sin2x}=\frac{1}{2\cos^2x} \).
2. \( \int\mathrm{cosec}\,2x\,dx=\frac12\log\tan x \), so IF \( =\frac{1}{\sqrt{\tan x}} \).
3. \( \frac{y}{\sqrt{\tan x}}=\int\frac{\sec^2x}{2\sqrt{\tan x}}dx=\sqrt{\tan x}+c \).
4. So \( y=\tan x+c\sqrt{\tan x} \).

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.

Answer: (c) \( u=(\log z)^{-1} \)

Explanation:
1. Divide by z and put \( w=\log z \): \( w'+\frac wx=\frac{w^2}{x^2} \).
2. This is Bernoulli with n = 2.
3. Dividing by \( w^2 \) and putting \( u=w^{-1} \) gives \( u'-\frac ux=-\frac{1}{x^2} \), which is linear.

Teacher's Note:
1. Spot the power of y on the right side to recognise a Bernoulli equation.
2. The same idea works for x as the dependent variable.

Answer: (a) \( (x-y)^2(1+y_1^2)=(x+yy_1)^2 \)

Explanation:
1. The circles are \( (x-a)^2+(y-a)^2=a^2 \).
2. Differentiating, \( (x-a)+(y-a)y_1=0 \), so \( a=\frac{x+yy_1}{1+y_1} \).
3. Substituting this value of a into the circle equation and simplifying gives \( (x-y)^2(1+y_1^2)=(x+yy_1)^2 \).

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: (d) \( y\left(\frac{dy}{dx}\right)^2+2x\frac{dy}{dx}-y=0 \)

Explanation:
1. With focus at the origin and axis along the x-axis the parabola is \( y^2=4a(x+a) \).
2. Differentiating, \( yy_1=2a \).
3. Substituting: \( y^2=2yy_1x+y^2y_1^2 \).
4. Divide by y: \( yy_1^2+2xy_1-y=0 \).

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: (a) \( x\left[yy_2+y_1^2\right]=yy_1 \)

Explanation:
1. The family is \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \).
2. Differentiating: \( \frac{x}{a^2}+\frac{yy_1}{b^2}=0 \), so \( \frac{yy_1}{x}=-\frac{b^2}{a^2} \).
3. Differentiate again: \( \frac{x(yy_2+y_1^2)-yy_1}{x^2}=0 \).
4. So \( x(yy_2+y_1^2)=yy_1 \).

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: (a) \( x^2y_2-4xy_1+6y=0 \)

Explanation:
1. \( y_1=3ax^2+2bx \) and \( y_2=6ax+2b \).
2. \( x^2y_2-4xy_1+6y=(6-12+6)ax^3+(2-8+6)bx^2=0 \).
3. So \( x^2y_2-4xy_1+6y=0 \).

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) \( \frac{dy}{dx}=\frac{x^2-c^2}{x^2+c^2} \)

Explanation:
1. The hyperbola has slope \( m_1=-\frac{c^2}{x^2} \).
2. \( \tan45^\circ=\frac{m-m_1}{1+mm_1}=1 \) gives \( m(1-m_1)=1+m_1 \).
3. \( m=\frac{1+m_1}{1-m_1}=\frac{x^2-c^2}{x^2+c^2} \).

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: (b) \( 8xy_1^3=12yy_1^2+27x \)

Explanation:
1. Differentiate: \( 3x^2=2c(y+c)y_1 \).
2. Dividing the original by this: \( \frac{x}{3}=\frac{y+c}{2y_1} \) gives \( y+c=\frac{2xy_1}{3} \) and \( c=\frac{9x}{4y_1^2} \).
3. Substituting c: \( y=\frac{2xy_1}{3}-\frac{9x}{4y_1^2} \).
4. Clear fractions: \( 8xy_1^3=12yy_1^2+27x \).

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: (b) \( y_4+2y_2+y=0 \)

Explanation:
1. The roots are \( \pm i \), each repeated twice.
2. \( (m^2+1)^2=m^4+2m^2+1 \).
3. So \( y_4+2y_2+y=0 \).

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: (b) \( x^2y_2-2xy_1+2y=0 \)

Explanation:
1. \( y_1=2ax+b \), \( y_2=2a \).
2. \( b=y_1-xy_2 \).
3. \( y=\frac{x^2y_2}{2}+x(y_1-xy_2)=xy_1-\frac{x^2y_2}{2} \).
4. So \( x^2y_2-2xy_1+2y=0 \).

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: (c) \( 6xy=3x^3+29x-6 \)

Explanation:
1. \( y=\frac{x^2}{2}-\frac1x+c \).
2. At (3, 9): \( 9=\frac92-\frac13+c \), so \( c=\frac{29}{6} \).
3. Multiply by 6x: \( 6xy=3x^3-6+29x \).

Teacher's Note:
1. Tangent and normal intercepts are expressed through y and dy/dx.
2. Setting up the equation correctly is most of the work.

Answer: (a) \( y=\frac12x^2\log x-\frac34x^2+C_1x+C_2 \)

Explanation:
1. \( \frac{dy}{dx}=\int\log x\,dx=x\log x-x+C_1 \).
2. \( y=\int(x\log x-x)dx+C_1x+C_2 \).
3. \( y=\frac{x^2}{2}\log x-\frac{x^2}{4}-\frac{x^2}{2}+C_1x+C_2 \).
4. So \( y=\frac12x^2\log x-\frac34x^2+C_1x+C_2 \).

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: (c) \( (1+x^2)^{3/2}+(1+y^2)^{3/2}=c \)

Explanation:
1. \( \int x\sqrt{1+x^2}dx=\frac13(1+x^2)^{3/2} \).
2. Similarly for y.
3. So \( (1+x^2)^{3/2}+(1+y^2)^{3/2}=c \).

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) \( x=cye^x \)

Explanation:
1. Divide by xy: \( \frac{dx}{x}-\frac{dy}{y}=dx \).
2. \( \log x-\log y=x+k \).
3. \( \frac{x}{y}=ce^x \), i.e. \( x=cye^x \).

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.