JEE Mathematics Differential Equations MCQs Set 15

Answer: (a) \( \log\left[(x+y)^2-ab\right]=\frac{2}{b-a}[x-y]+k \)

Explanation:
1. Put \( v=x+y \): \( \frac{dv}{dx}=\frac{2(v^2-ab)}{(v+b)(v-a)} \).
2. \( \frac{(v+b)(v-a)}{v^2-ab}dv=2\,dx \), i.e. \( \left[1+\frac{(b-a)v}{v^2-ab}\right]dv=2\,dx \).
3. \( v+\frac{b-a}{2}\log(v^2-ab)=2x+c \).
4. Hence \( \log[(x+y)^2-ab]=\frac{2}{b-a}(x-y)+k \).

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: (c) \( a=0,\ b\ne0 \)

Explanation:
1. \( (by+k)dy=(ax+h)dx \).
2. \( \frac{by^2}{2}+ky=\frac{ax^2}{2}+hx+c \).
3. A parabola arises when exactly one of the squared terms is present, i.e. \( a=0,\ b\ne0 \) (or the reverse).

Teacher's Note:
1. Compare with the standard forms in the table of solutions.
2. A quick check is to differentiate the answer.

Answer: (b) a rectangular hyperbola

Explanation:
1. The normal meets the x-axis at \( (x+yy_1,0) \).
2. Equal sides from the point mean its foot lies at the midpoint of the base: \( x+yy_1=2x \).
3. \( y\,y_1=x \), so \( y^2-x^2=c \), a rectangular 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: (c) a rectangular hyperbola

Explanation:
1. \( \frac{dy}{dx}=\frac xy \), so \( y\,dy=x\,dx \).
2. \( y^2-x^2=c \), which is a rectangular hyperbola.

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) \( \frac{x^2}{y}+x^3y^2=c \)

Explanation:
1. Let \( F=\frac{x^2}{y}+x^3y^2 \).
2. \( F_x=\frac{2x}{y}+3x^2y^2 \) and \( F_y=-\frac{x^2}{y^2}+2x^3y \).
3. \( -\frac{F_x}{F_y}=\frac{2xy+3x^2y^4}{x^2-2x^3y^3} \), which is the given right side.
4. So \( \frac{x^2}{y}+x^3y^2=c \).

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) \( \frac yx-\frac{1}{2(x^2+y^2)}=c \)

Explanation:
1. The right side is \( \frac{(x^2+y^2)^2}{x^2} \).
2. Cross-multiplying: \( x\,dx+y\,dy=\frac{(x^2+y^2)^2}{x^2}(y\,dx-x\,dy)=-(x^2+y^2)^2d\!\left(\frac yx\right) \).
3. With \( r^2=x^2+y^2 \): \( \frac{r\,dr}{r^4}=-d\!\left(\frac yx\right) \).
4. \( -\frac{1}{2r^2}=-\frac yx+c \), so \( \frac yx-\frac{1}{2(x^2+y^2)}=c \).

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) \( \sqrt{1-x^2} \)

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

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) \( -\frac{\sin3x}{9}+e^x+\frac{x^4}{12}+\frac13x-1 \)

Explanation:
1. \( y'=-\frac{\cos3x}{3}+e^x+\frac{x^3}{3}+C_1 \); \( y'(0)=1 \) gives \( C_1=\frac13 \).
2. \( y=-\frac{\sin3x}{9}+e^x+\frac{x^4}{12}+\frac x3+C_2 \); \( y(0)=0 \) gives \( C_2=-1 \).

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) 50 cm/s

Explanation:
1. Acceleration is zero at \( t=20 \).
2. \( v=\int_0^t\left(5-\frac s4\right)ds=5t-\frac{t^2}{8} \).
3. At t = 20: \( v=100-50=50 \) cm/s.

Teacher's Note:
1. Rate problems are solved by writing dQ/dt equal to a constant times Q or a similar relation.
2. Fix the constants using the initial conditions.

Answer: (b) \( \sqrt{\frac53} \)

Explanation:
1. \( 2x\,dx=3y\,dy \), so \( x^2-\frac32y^2=k \).
2. Write \( \frac{x^2}{k}-\frac{y^2}{2k/3}=1 \): \( a^2=k,\ b^2=\frac{2k}{3} \).
3. \( e^2=1+\frac{b^2}{a^2}=\frac53 \), so \( e=\sqrt{\frac53} \).

Teacher's Note:
1. Compare with the standard forms in the table of solutions.
2. A quick check is to differentiate the answer.

Answer: (a) \( \frac56 \)

Explanation:
1. \( y=x^2+x+c \); through (1, 2): c = 0.
2. The curve meets the x-axis at x = 0 and x = -1; the region to the right of the y-axis up to x = 1 is \( \int_0^1(x^2+x)dx \).
3. \( =\frac13+\frac12=\frac56 \).

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) 1

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

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) 80 years

Explanation:
1. \( P=P_0e^{kt} \) with \( e^{50k}=2 \), so \( k=\frac{\ln2}{50} \).
2. \( e^{kt}=3 \) gives \( t=\frac{50\ln3}{\ln2}\approx79.2 \).
3. The nearest option is 80 years.

Teacher's Note:
1. Rate problems are solved by writing dQ/dt equal to a constant times Q or a similar relation.
2. Fix the constants using the initial conditions.

Answer: (b) \( y=c-\log|\sin x+\cos x| \)

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

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) \( y=\frac{x^3}{6}+c \)

Explanation:
1. Slope of the hyperbola: \( -\frac{2}{x^2} \).
2. The orthogonal slope is \( \frac{x^2}{2} \).
3. \( y=\frac{x^3}{6}+c \).

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: (a) \( f\left(\frac{y^2}{x^2}\right)=cx^2 \)

Explanation:
1. Put \( u=y^2=wx^2 \). Then \( \frac12u'=yy_1 \) gives \( xw+\frac{x^2w'}{2}=xw+x\frac{f(w)}{f'(w)} \).
2. \( \frac{f'(w)}{f(w)}dw=\frac{2\,dx}{x} \).
3. \( \log f(w)=2\log x+\log c \), so \( f\left(\frac{y^2}{x^2}\right)=cx^2 \).

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: (c) 1

Explanation:
1. The slope at a point is the value of the right side there.
2. At (1, 0): \( \frac{1+0}{1-0}=1 \).

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: (d) \( \frac{2x^2-1}{x(1-x^2)} \)

Explanation:
1. Divide by \( x(1-x^2) \): \( \frac{dy}{dx}+\frac{2x^2-1}{x(1-x^2)}y=\frac{ax^2}{1-x^2} \).
2. The coefficient of y is p.

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: (d) hyperbola

Explanation:
1. M = (x, 0) and N = \( (0,\ y-xy_1) \).
2. \( \frac12x(y-xy_1)=k \), so \( y_1-\frac yx=-\frac{2k}{x^2} \).
3. \( \frac{d}{dx}\left(\frac yx\right)=-\frac{2k}{x^3} \), so \( \frac yx=\frac{k}{x^2}+c \).
4. \( xy=k+cx^2 \) is a hyperbola.

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: (c) ellipse and straight line

Explanation:
1. \( y'+\frac{x}{1-x^2}y=\frac{ax}{1-x^2} \).
2. IF \( =(1-x^2)^{-1/2} \).
3. \( \frac{y}{\sqrt{1-x^2}}=\int\frac{ax}{(1-x^2)^{3/2}}dx=\frac{a}{\sqrt{1-x^2}}+c \).
4. \( y=a+c\sqrt{1-x^2} \), so \( x^2+\frac{(y-a)^2}{c^2}=1 \) (an ellipse), and for c = 0 the line y = a.

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: (b) \( \sin y^2=(x+1)^2\left[\frac{(x+1)^2}{2}+c\right] \)

Explanation:
1. Put \( u=\sin y^2 \): \( u'=2y\cos y^2\,y' \).
2. \( u'-\frac{2}{x+1}u=(x+1)^3 \).
3. IF \( (x+1)^{-2} \): \( \frac{u}{(x+1)^2}=\int(x+1)dx=\frac{(x+1)^2}{2}+c \).

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: (b) \( 3y^2=2x^2e^{1/x^3}+cx^2 \)

Explanation:
1. Put \( u=y^2 \): \( \frac{u'}{2}-\frac ux=-\frac{e^{1/x^3}}{x^2} \).
2. \( u'-\frac{2u}{x}=-\frac{2e^{1/x^3}}{x^2} \); IF \( x^{-2} \).
3. \( \frac{u}{x^2}=-2\int x^{-4}e^{x^{-3}}dx=\frac23e^{x^{-3}}+c \).
4. \( 3y^2=2x^2e^{1/x^3}+cx^2 \).

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.