JEE Mathematics Differential Equations MCQs Set 12

Answer: (d) hyperbola

Explanation:
1. \( y\frac{dy}{dx}=x \), so \( y\,dy=x\,dx \).
2. \( y^2=x^2+c \).
3. Using (4, 2): \( 4=16+c \), c = -12.
4. \( x^2-y^2=12 \) is a hyperbola.

Teacher's Note:
1. Translate the geometric condition (slope, intercept, area or distance) into a differential equation, then solve.
2. Use the given point to fix the constant.

Answer: (b) \( (y^2-x^2)dx-2xy\,dy=0 \)

Explanation:
1. The circles are \( x^2+y^2+2gx=0 \).
2. Differentiating, \( x+yy_1+g=0 \), so \( 2g=-2(x+yy_1) \).
3. Substituting: \( x^2+y^2-2x(x+yy_1)=0 \), i.e. \( y^2-x^2-2xyy_1=0 \).
4. So \( (y^2-x^2)dx-2xy\,dy=0 \).

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) \( \frac{dr}{dt}=-k \)

Explanation:
1. \( \frac{dV}{dt}=-k'(4\pi r^2) \) with \( V=\frac43\pi r^3 \).
2. \( 4\pi r^2\frac{dr}{dt}=-k'\cdot4\pi r^2 \).
3. So \( \frac{dr}{dt}=-k \).

Teacher's Note:
1. Read the verbal rate as a differential equation, solve it and substitute the given values.

Answer: (d) \( y_4=y \)

Explanation:
1. Each of the four functions satisfies \( f_4=f \).
2. So \( y_4=y \).

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) \( y=x^4y_1^2-xy_1 \)

Explanation:
1. Differentiating, \( y_1=-\frac{c}{x^2} \), so \( c=-x^2y_1 \).
2. Substituting: \( y=x^4y_1^2-xy_1 \).

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

Explanation:
1. A non-vertical line is \( y=mx+c \).
2. Differentiating twice: \( \frac{d^2y}{dx^2}=0 \).

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: (d) \( y_2+y=2\cos x \)

Explanation:
1. Differentiate twice: \( y_2=-a\cos x-b\sin x+2\cos x-x\sin x \).
2. Adding y: \( y_2+y=2\cos x \).

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) \( xy_2+2y_1=xy \)

Explanation:
1. Let \( u=xy \). Then \( u''=u \).
2. \( u''=xy_2+2y_1 \).
3. So \( xy_2+2y_1=xy \).

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) \( (1-x^2)y_2-xy_1=2 \)

Explanation:
1. \( y_1=\frac{2\sin^{-1}x-A}{\sqrt{1-x^2}} \).
2. So \( \sqrt{1-x^2}\,y_1=2\sin^{-1}x-A \).
3. Differentiating: \( \sqrt{1-x^2}\,y_2-\frac{x}{\sqrt{1-x^2}}y_1=\frac{2}{\sqrt{1-x^2}} \).
4. Multiply by \( \sqrt{1-x^2} \): \( (1-x^2)y_2-xy_1=2 \).

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: (b) circles with centre on the x-axis

Explanation:
1. \( y\,dy+x\,dx=A\,dx \).
2. \( \frac{y^2}{2}+\frac{x^2}{2}=Ax+c \), i.e. \( (x-A)^2+y^2=A^2+2c \).
3. These are circles with centre (A, 0) on the x-axis.

Teacher's Note:
1. Translate the geometric condition (slope, intercept, area or distance) into a differential equation, then solve.
2. Use the given point to fix the constant.

Answer: (d) \( y=k(e^x-1) \)

Explanation:
1. \( \frac{y_2}{y_1}=1 \) gives \( y_1=c_1e^x \).
2. \( y=c_1e^x+c_2 \).
3. Through the origin: \( c_2=-c_1 \).
4. So \( y=k(e^x-1) \).

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: (b) \( 2y=x^2\left[\log x-\frac12\right]+c \)

Explanation:
1. \( y=\int x\log x\,dx=\frac{x^2}{2}\log x-\int\frac{x}{2}dx \).
2. \( y=\frac{x^2}{2}\log x-\frac{x^2}{4}+c \).
3. So \( 2y=x^2\left[\log x-\frac12\right]+c' \).

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) \( y\sin^{-1}x=c \)

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

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) \( (x^2+y^2)e^{2y}=c \)

Explanation:
1. Divide by \( x^2+y^2 \): \( \frac{x\,dx+y\,dy}{x^2+y^2}+dy=0 \).
2. \( \frac12\log(x^2+y^2)+y=k \).
3. So \( (x^2+y^2)e^{2y}=c \).

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) \( x^2-1=c(1+y^2) \)

Explanation:
1. Rearrange: \( x(1+y^2)dx=y(x^2-1)dy \).
2. \( \frac{x\,dx}{x^2-1}=\frac{y\,dy}{1+y^2} \).
3. \( \frac12\log(x^2-1)=\frac12\log(1+y^2)+\frac12\log c \).
4. So \( x^2-1=c(1+y^2) \).

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) \( x=-(y+2)+ce^y \)

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

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) \( x+c=\tan\left(\frac{x+y}{2}\right) \)

Explanation:
1. \( \frac{dy}{dx}=\cos(x+y) \).
2. Put \( v=x+y \): \( \frac{dv}{dx}=1+\cos v=2\cos^2\frac v2 \).
3. \( \frac12\sec^2\frac v2\,dv=dx \).
4. \( \tan\frac v2=x+c \).

Teacher's Note:
1. When the right side depends on a combination like x + y, put that combination equal to v.
2. Then dy/dx = dv/dx - 1.

Answer: (a) \( 2y=c+a\log\left(\frac{x-y-a}{x-y+a}\right) \)

Explanation:
1. Put \( v=x-y \): \( 1-\frac{dv}{dx}=\frac{a^2}{v^2} \), so \( \frac{v^2}{v^2-a^2}dv=dx \).
2. \( v+\frac a2\log\frac{v-a}{v+a}=x+k \).
3. Replace v: \( -y+\frac a2\log\frac{x-y-a}{x-y+a}=k \), so \( 2y=c+a\log\frac{x-y-a}{x-y+a} \).

Teacher's Note:
1. A substitution such as v = x + y or v = y - x turns the equation into a separable one.
2. Always return to x and y at the end.

Answer: (a) \( e^{-(x+y)}+x+c=0 \)

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

Teacher's Note:
1. When the right side depends on a combination like x + y, put that combination equal to v.
2. Then dy/dx = dv/dx - 1.

Answer: (a) \( x^2-y^2=2(ax-by) \)

Explanation:
1. \( (y-b)dy=(x-a)dx \).
2. \( \frac{y^2}{2}-by=\frac{x^2}{2}-ax+c \).
3. Through the origin c = 0.
4. So \( x^2-y^2=2(ax-by) \).

Teacher's Note:
1. Translate the geometric condition (slope, intercept, area or distance) into a differential equation, then solve.
2. Use the given point to fix the constant.

Answer: (c) \( y=ce^{x/k} \)

Explanation:
1. Subtangent is \( \frac{y}{y_1}=k \).
2. \( \frac{dy}{y}=\frac{dx}{k} \).
3. \( y=ce^{x/k} \).

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) \( e^{2x}(1-x+y)=1+x-y \)

Explanation:
1. Put \( v=x-y \): \( \frac{dv}{dx}=1-v^2 \).
2. \( \frac12\log\frac{1+v}{1-v}=x+c \); at the origin c = 0.
3. \( 1+v=e^{2x}(1-v) \), i.e. \( e^{2x}(1-x+y)=1+x-y \).

Teacher's Note:
1. When the right side depends on a combination like x + y, put that combination equal to v.
2. Then dy/dx = dv/dx - 1.

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

Explanation:
1. Separate: \( xe^x\,dx+\frac{y\,dy}{\sqrt{1-y^2}}=0 \).
2. \( \int xe^xdx=(x-1)e^x \) and \( \int\frac{y\,dy}{\sqrt{1-y^2}}=-\sqrt{1-y^2} \).
3. So \( (x-1)e^x-\sqrt{1-y^2}=c \).

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.