JEE Mathematics Differential Equations MCQs Set 20

Answer: (a) 2, 2

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

Teacher's Note:
1. Degree is defined only after the equation is free of radicals and fractions in the derivatives.
2. The power of the highest-order derivative gives the degree.

Answer: (b) 3

Explanation:
1. Here \( C_1+C_2 \) acts as one constant, and \( C_4e^{C_5} \) also acts as one constant.
2. The essential constants are therefore \( C_1+C_2,\ C_3,\ C_4e^{C_5} \), i.e. three.
3. The order equals the number of essential arbitrary constants, so it is 3.

Teacher's Note:
1. Combine constants that appear only as a sum or product.
2. Order = number of independent arbitrary constants.

Answer: (a) 1, 2

Explanation:
1. The tangent at \( (2t,t^2) \) is \( y=tx-t^2 \), where the slope is \( t=y' \).
2. Eliminating the parameter gives \( y=xy'-(y')^2 \).
3. It contains only the first derivative, with power 2: order 1, degree 2.

Teacher's Note:
1. Use the point-parameter form of the tangent to a parabola.
2. One parameter means a first-order equation.

Answer: (c) p = q

Explanation:
1. The highest-order derivative is \( d^2y/dx^2 \), so p = 2.
2. It appears squared, so q = 2.
3. Hence p = q.

Teacher's Note:
1. The equation is already polynomial in the derivatives.
2. Compare the order with the power of the highest derivative.

Answer: (c) three

Explanation:
1. Differentiate: \( y'=A \).
2. Substitute: \( y=xy'+(y')^3 \).
3. The first derivative appears to the power 3, so the degree is 3.

Teacher's Note:
1. Eliminate A using the derivative.
2. Look at the highest power of y' in the resulting equation.

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

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 \( x-h=y'\frac{1+y'^2}{y''} \).
3. Substitute in the circle: \( \frac{(1+y'^2)^3}{y''^2}=a^2 \), i.e. \( (1+y'^2)^3=a^2y''^2 \).

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

Answer: (b) 3

Explanation:
1. Such conics are \( ax^2+2hxy+by^2=1 \).
2. There are three independent constants a, h, b.
3. So the order is 3.

Teacher's Note:
1. Count the independent constants in the general central conic.
2. Order = number of constants.

Answer: (d) \( \left(y-x\frac{dy}{dx}\right)^2=1+\left(\frac{dy}{dx}\right)^2 \)

Explanation:
1. A line is \( y=mx+c \) with distance \( \frac{|c|}{\sqrt{1+m^2}}=1 \), so \( c^2=1+m^2 \).
2. Here \( m=y' \) and \( c=y-xy' \).
3. Therefore \( (y-xy')^2=1+y'^2 \).

Teacher's Note:
1. Use the point-to-line distance formula.
2. Write c as y - m x with m = y'.

Answer: (d) \( \frac{e^6+9}{2} \)

Explanation:
1. Separate: \( e^{2y}dy=dx \), so \( \frac{e^{2y}}2=x+c \).
2. At x = 5, y = 0: \( \frac12=5+c\Rightarrow c=-\frac92 \).
3. For y = 3: \( x=\frac{e^6}{2}+\frac92=\frac{e^6+9}2 \).

Teacher's Note:
1. Separate the variables and integrate.
2. Use the point (5, 0) to find the constant.

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

Explanation:
1. The equation gives \( \phi(x)=Ce^x \).
2. From \( \phi(1)=2 \): \( C=2/e \).
3. Hence \( \phi(3)=\frac2e e^3=2e^2 \).

Teacher's Note:
1. The function equals its own derivative.
2. Fix the constant using the value at x = 1.

Answer: (a) \( e^{(1+x)^2/2}-1 \)

Explanation:
1. Factorise: \( y'=(1+x)(1+y) \).
2. Separate: \( \frac{dy}{1+y}=(1+x)dx \), so \( \ln(1+y)=\frac{(1+x)^2}{2}+c \).
3. At x = -1, y = 0: c = 0. So \( y=e^{(1+x)^2/2}-1 \).

Teacher's Note:
1. 1 + x + y + xy factorises as (1 + x)(1 + y).
2. Check the option against y(-1) = 0.

Answer: (b) 1

Explanation:
1. Separate: \( \frac{dy}{y(1-y)}=dx \), so \( \ln\left|\frac{y}{1-y}\right|=x+c \).
2. With y(0) = 2: \( \frac{y}{y-1}=2e^{x} \), so \( y=\frac{2e^x}{2e^x-1} \).
3. As x tends to infinity, y tends to 1.

Teacher's Note:
1. Use partial fractions on 1/(y(1-y)).
2. Divide numerator and denominator by e^x for the limit.

Answer: (b) \( \sin^{-1}x+\sin^{-1}y=C \)

Explanation:
1. Separate: \( \frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}} \).
2. Integrate: \( \sin^{-1}y=-\sin^{-1}x+C \).
3. So \( \sin^{-1}x+\sin^{-1}y=C \).

Teacher's Note:
1. Both sides are standard arcsine integrals.
2. Mind the minus sign when separating.

Answer: (d) \( -\frac53 \)

Explanation:
1. The slope at (1, 2) is the value of the right side there.
2. Substitute: \( \frac{1+4}{1-4}=-\frac53 \).

Teacher's Note:
1. The slope of an integral curve at a point is simply dy/dx there.
2. No need to solve the equation.

Answer: (b) a straight line through the origin

Explanation:
1. Divide by \( xy \): \( \frac{dy}{y}=\frac{dx}{x} \).
2. Integrate: \( \ln y=\ln x+\ln c \), so \( y=cx \).
3. This is a family of lines through the origin.

Teacher's Note:
1. Separate the variables.
2. y = cx is linear and passes through (0, 0).

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

Explanation:
1. The condition gives \( \frac{dy}{dx}=\frac1{2y} \), i.e. \( 2y\,dy=dx \).
2. Integrate: \( y^2=x+c \).
3. Through (4, 3): \( 9=4+c \), so c = 5 and \( y^2=x+5 \).

Teacher's Note:
1. Translate the word statement into dy/dx = 1/(2y).
2. Use the point to fix c.

Answer: (c) \( \sqrt{x^2-y^2}=Ae^{4\sin^{-1}(y/x)} \)

Explanation:
1. Put \( y=vx \): \( x\,dy-y\,dx=x^2dv \) and \( x\,dx-y\,dy=x(1-v^2)dx-vx^2dv \).
2. The equation becomes \( \frac{dx}{x}=\frac{v\,dv}{1-v^2}+\frac{4\,dv}{\sqrt{1-v^2}} \).
3. Integrate: \( \ln x=-\tfrac12\ln(1-v^2)+4\sin^{-1}v+c \), i.e. \( x\sqrt{1-v^2}=Ae^{4\sin^{-1}v} \).
4. Since \( x\sqrt{1-v^2}=\sqrt{x^2-y^2} \); this is the stated result.

Teacher's Note:
1. The equation is homogeneous: use y = vx.
2. Integrate the arcsine term separately.

Answer: (d) \( y^2=2ax+b \)

Explanation:
1. For P(x, y) the normal meets the x-axis at \( x+yy' \).
2. So \( P'N=|yy'|=a \) (taking the constant with sign).
3. Then \( y\,dy=a\,dx \) and \( y^2=2ax+b \).

Teacher's Note:
1. The x-intercept of the normal is x + y y'.
2. P'N is the sub-normal.

Answer: (b) \( x^2-y^2+5xy-3y=0 \)

Explanation:
1. Cross-multiply: \( (2y-5x+3)dy-(2x+5y)dx=0 \).
2. This is exact: it equals \( d(y^2+3y-5xy-x^2)=0 \).
3. So \( y^2+3y-5xy-x^2=c \); y(0) = 0 gives c = 0, i.e. \( x^2-y^2+5xy-3y=0 \).

Teacher's Note:
1. Rearrange into M dx + N dy = 0 and test for exactness.
2. Use the condition y(0) = 0.

Answer: (d) \( 4y=x+\ln|3x+4y|+C \)

Explanation:
1. Put \( u=3x+4y \): \( \frac{du}{dx}=3+4\cdot\frac{u+3}{4u-4}=\frac{4u}{u-1} \).
2. Separate: \( \left(1-\frac1u\right)du=4\,dx \), so \( u-\ln|u|=4x+c \).
3. Substituting back: \( 4y=x+\ln|3x+4y|+C \).

Teacher's Note:
1. The coefficients 3, 4 and 12, 16 are proportional.
2. Substituting u = 3x + 4y makes the variables separable.

Answer: (c) \( v=ce^{-kt/m}-\frac{mg}{k} \)

Explanation:
1. Integrating factor: \( e^{kt/m} \).
2. Then \( \frac{d}{dt}(ve^{kt/m})=-ge^{kt/m} \), so \( ve^{kt/m}=-\frac{mg}{k}e^{kt/m}+c \).
3. Hence \( v=ce^{-kt/m}-\frac{mg}{k} \).

Teacher's Note:
1. This is linear with constant coefficients.
2. Multiply by the integrating factor e^(kt/m).

Answer: (a) \( y^4x=\frac{x^5}{5}+C \)

Explanation:
1. Put \( w=y^4 \), so \( w'=4y^3y' \).
2. Then \( w'+\frac wx=x^3 \), with integrating factor x: \( (xw)'=x^4 \).
3. So \( xw=\frac{x^5}5+C \), i.e. \( xy^4=\frac{x^5}{5}+C \).

Teacher's Note:
1. The substitution w = y^4 makes it linear.
2. Integrating factor is x.