JEE Mathematics Differential Equations MCQs Set 09

Answer: (d) \( A=-13,\ B=-12 \)

Explanation:
1. The exponents 4 and -1 must satisfy \( m^3+Am+B=0 \).
2. \( 64+4A+B=0 \) and \( -1-A+B=0 \).
3. Subtracting: \( 65+5A=0 \), so \( A=-13 \) and \( B=-12 \).

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) \( y_3-6y_2+11y_1-6y=0 \)

Explanation:
1. The exponents are 1, 2, 3.
2. \( (m-1)(m-2)(m-3)=m^3-6m^2+11m-6 \).
3. Hence \( y_3-6y_2+11y_1-6y=0 \).

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) \( y_2-6y_1+10y=0 \)

Explanation:
1. The roots are \( 3\pm i \).
2. The quadratic is \( m^2-6m+10=0 \).
3. So \( y_2-6y_1+10y=0 \).

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) \( \frac{dy}{dx}=y\cot x \)

Explanation:
1. Differentiating, \( y_1=c\cos x \).
2. Since \( c=\frac{y}{\sin x} \), \( y_1=y\cot x \).
3. This is first order, the lowest possible.

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) \( y_3=0 \)

Explanation:
1. Differentiating three times: \( y_1=2ax+b,\ y_2=2a,\ y_3=0 \).
2. So \( y_3=0 \).

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) \( xy_2+2y_1=xy+2-x^2 \)

Explanation:
1. Differentiate twice: \( y+xy_1=ae^x-be^{-x}+2x \) and \( 2y_1+xy_2=ae^x+be^{-x}+2 \).
2. The right side equals \( xy-x^2+2 \).
3. So \( xy_2+2y_1=xy+2-x^2 \).

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: (a) \( y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2+1=0 \)

Explanation:
1. Differentiate: \( x+yy_1+\lambda=0 \).
2. Differentiate again: \( 1+y_1^2+yy_2=0 \).
3. The constants have been eliminated.

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) \( \frac{1-x}{1-y}=c \)

Explanation:
1. \( \frac{dy}{1-y}=\frac{dx}{1-x} \).
2. Integrating, \( -\log|1-y|=-\log|1-x|+k \).
3. Hence \( \frac{1-x}{1-y}=c \).

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: (a) \( y=a\cos x \)

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

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

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

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

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

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: (a) \( 3(y^2-x^2)=2(x^3+y^3)+c \)

Explanation:
1. \( (y-y^2)dy=(x+x^2)dx \).
2. \( \frac{y^2}{2}-\frac{y^3}{3}=\frac{x^2}{2}+\frac{x^3}{3}+k \).
3. Multiply by 6: \( 3y^2-2y^3=3x^2+2x^3+c \), which is option (a).

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) \( \frac{y^2}{2}=\log c(e^x+1) \)

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

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) \( y=cx \)

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

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) \( x+ye^{x^3}-cy=0 \)

Explanation:
1. Divide by \( y^2 \): \( \frac{y\,dx-x\,dy}{y^2}+3x^2e^{x^3}dx=0 \).
2. Since \( d\!\left(\frac xy\right)=\frac{y\,dx-x\,dy}{y^2} \) and \( d(e^{x^3})=3x^2e^{x^3}dx \), the equation is \( d\!\left(\frac xy+e^{x^3}\right)=0 \).
3. So \( \frac xy+e^{x^3}=c \), i.e. \( x+ye^{x^3}-cy=0 \).

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) \( x+y+1=xy \)

Explanation:
1. \( \tan^{-1}x+\tan^{-1}y=C \), so \( \frac{x+y}{1-xy}=k \).
2. At (0, -1): \( k=-1 \).
3. Then \( x+y=-(1-xy) \), i.e. \( x+y+1=xy \).

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) \( \tan y=c(1-e^x) \)

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

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) \( \tan(xy)-\cos x-\cos y=c \)

Explanation:
1. Regroup: \( \frac{y\,dx+x\,dy}{\cos^2(xy)}+\sin x\,dx+\sin y\,dy=0 \).
2. \( \sec^2(xy)\,d(xy)+\sin x\,dx+\sin y\,dy=0 \).
3. Integrating, \( \tan(xy)-\cos x-\cos y=c \).

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) \( x=ce^{kt} \)

Explanation:
1. \( \frac{dx}{dt}=kx \).
2. \( \frac{dx}{x}=k\,dt \).
3. \( \log x=kt+\log c \), so \( x=ce^{kt} \).

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

Explanation:
1. Put \( y=vx \): \( x^2(v\,dx+x\,dv)+vx(x+vx)dx=0 \).
2. \( x\,dv+(2v+v^2)dx=0 \), so \( \frac{dv}{v(v+2)}=-\frac{dx}{x} \).
3. \( \frac12\log\frac{v}{v+2}=-\log x+\log c \), giving \( \frac{y}{y+2x}=\frac{c^2}{x^2} \).
4. So \( x^2y=c^2(2x+y) \).

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: (a) \( \tan\left(\frac{y}{2x}\right)-cx=0 \)

Explanation:
1. Put \( y=vx \): \( v+x\frac{dv}{dx}=v+\sin v \).
2. \( \frac{dv}{\sin v}=\frac{dx}{x} \).
3. \( \log\tan\frac v2=\log x+\log c \).
4. So \( \tan\frac{y}{2x}=cx \).

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: (a) \( \tan^{-1}\left(\frac xy\right)+\log y+c=0 \)

Explanation:
1. Put \( x=vy \): \( y^2(v\,dy+y\,dv)+(v^2-v+1)y^2dy=0 \).
2. \( (v^2+1)dy+y\,dv=0 \).
3. \( \frac{dy}{y}+\frac{dv}{1+v^2}=0 \).
4. \( \log y+\tan^{-1}\frac xy+c=0 \).

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.