JEE Mathematics Differential Equations MCQs Set 14

Answer: (c) \( x^3+3xy+y^3=c \)

Explanation:
1. Regroup: \( x^2dx+(y\,dx+x\,dy)+y^2dy=0 \).
2. \( \frac{x^3}{3}+xy+\frac{y^3}{3}=k \).
3. So \( x^3+3xy+y^3=c \).

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) \( be^{ax}+ae^{-by}=k \)

Explanation:
1. \( \frac{dy}{dx}=e^{ax}e^{by} \).
2. \( e^{-by}dy=e^{ax}dx \).
3. \( -\frac{e^{-by}}{b}=\frac{e^{ax}}{a}+c \).
4. Multiply by -ab: \( be^{ax}+ae^{-by}=k \).

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) \( \sin(y-x)=ce^{x^2/2} \)

Explanation:
1. Put \( v=y-x \): \( \frac{dv}{dx}=x\tan v \).
2. \( \cot v\,dv=x\,dx \).
3. \( \log\sin v=\frac{x^2}{2}+\log c \).
4. So \( \sin(y-x)=ce^{x^2/2} \).

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) \( e^{3x}+3e^{-y}=4 \)

Explanation:
1. \( e^{-y}dy=e^{3x}dx \).
2. \( -e^{-y}=\frac{e^{3x}}{3}+c \).
3. At (0, 0): \( -1=\frac13+c \), so \( c=-\frac43 \).
4. \( e^{3x}+3e^{-y}=4 \).

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: (d) \( y\tan y-x\tan x+\log\left(\frac{\cos y}{\cos x}\right)=c \)

Explanation:
1. \( x\sec^2x\,dx=y\sec^2y\,dy \).
2. \( \int x\sec^2x\,dx=x\tan x+\log\cos x \).
3. So \( x\tan x+\log\cos x=y\tan y+\log\cos y+k \).
4. Rearranged: \( y\tan y-x\tan x+\log\frac{\cos y}{\cos x}=c \).

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) \( \log\left(\frac{x}{1+\sqrt{1+x^2}}\right)+\sqrt{1+x^2}+\sqrt{1+y^2}=c \)

Explanation:
1. The root is \( \sqrt{(1+x^2)(1+y^2)} \).
2. Separate: \( \frac{y\,dy}{\sqrt{1+y^2}}=-\frac{\sqrt{1+x^2}}{x}dx \).
3. \( \int\frac{\sqrt{1+x^2}}{x}dx=\sqrt{1+x^2}+\log\frac{x}{1+\sqrt{1+x^2}} \).
4. So \( \sqrt{1+y^2}+\sqrt{1+x^2}+\log\frac{x}{1+\sqrt{1+x^2}}=c \).

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.

Answer: (a) \( (y+2)^2=c\,e^{(x-3)^2} \)

Explanation:
1. Factorise: \( xy+2x-3y-6=(x-3)(y+2) \).
2. \( \frac{dy}{y+2}=(x-3)dx \).
3. \( \log(y+2)=\frac{(x-3)^2}{2}+k \).
4. So \( (y+2)^2=ce^{(x-3)^2} \).

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

Explanation:
1. \( \frac{dx}{x}=\frac{y\,dy}{1+y^2} \).
2. \( \log x=\frac12\log(1+y^2)+\log c \), so \( x^2=c^2(1+y^2) \).
3. Through (1, 0): \( c^2=1 \).
4. \( x^2-y^2=1 \).

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

Explanation:
1. Separate: \( xe^{x^2}dx=ye^{-y}dy \).
2. \( \frac12e^{x^2}=-(y+1)e^{-y}+k \).
3. So \( e^{x^2}+2(y+1)e^{-y}=c \).

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) \( x\,dv+(v^2+2v)dx=0 \)

Explanation:
1. \( x^2(v\,dx+x\,dv)+vx(x+vx)dx=0 \).
2. Divide by \( x^2 \): \( x\,dv+(v+v+v^2)dx=0 \).
3. So \( x\,dv+(v^2+2v)dx=0 \).

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.

Answer: (a) \( x^{-2} \)

Explanation:
1. Rewrite: \( \frac{dy}{dx}-\frac2xy=2x^3+x \).
2. \( \int-\frac2xdx=-2\log x \).
3. IF \( =x^{-2} \).

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: (a) \( e^{1/y} \)

Explanation:
1. Invert: \( \frac{dx}{dy}=\frac{x}{y^2}+2y \), so \( \frac{dx}{dy}-\frac{x}{y^2}=2y \).
2. \( \int-\frac{1}{y^2}dy=\frac1y \).
3. IF \( =e^{1/y} \).

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: (b) \( \sec y \)

Explanation:
1. Invert: \( \frac{dx}{dy}+x\tan y=\sec y \).
2. \( \int\tan y\,dy=\log\sec y \).
3. IF \( =\sec y \).

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: (a) \( \frac{x}{2+e^y}=-\frac{1}{2+e^y}+c \)

Explanation:
1. Multiply by \( e^y \): \( (x+1)e^y\frac{dy}{dx}=2+e^y \).
2. \( \frac{e^ydy}{2+e^y}=\frac{dx}{x+1} \).
3. \( \log(2+e^y)=\log(x+1)+\log k \), so \( 2+e^y=k(x+1) \).
4. This is \( \frac{x}{2+e^y}=-\frac{1}{2+e^y}+\frac1k \), option (a). With the condition, k = 3.

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

Explanation:
1. Here \( m=-a \), so the right side is \( e^{-ax} \).
2. IF \( =e^{ax} \).
3. \( ye^{ax}=\int e^{ax}e^{-ax}dx=x+c \).

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: (c) \( 3\sin x+c \)

Explanation:
1. The left side is \( \frac{d}{dx}(x^3y) \).
2. \( x^3y=\int3\cos x\,dx=3\sin x+c \).

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: (b) \( y\cos x=e^x+c \)

Explanation:
1. IF \( =e^{-\int\tan x}=\cos x \).
2. \( y\cos x=\int e^x\sec x\cos x\,dx=\int e^xdx=e^x+c \).

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) \( y=-(x+1)+e^x \)

Explanation:
1. \( \frac{dy}{dx}-y=x \).
2. IF \( =e^{-x} \).
3. \( ye^{-x}=\int xe^{-x}dx=-(x+1)e^{-x}+c \).
4. Through the origin: \( 0=-1+c \), c = 1.
5. \( y=-(x+1)+e^x \).

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

Explanation:
1. Write \( dx+y(1+x^2)dx+x(1+x^2)dy=0 \).
2. Divide by \( 1+x^2 \): \( \frac{dx}{1+x^2}+d(xy)=0 \).
3. \( \tan^{-1}x+xy=c \), so \( xy=-\tan^{-1}x+c \).

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: (b) \( \frac1x=cy-y\log y \)

Explanation:
1. Divide by \( x^2 \) and put \( u=\frac1x \): \( u'-\frac uy=-1 \).
2. IF \( =\frac1y \): \( \frac uy=-\log y+c \).
3. So \( \frac1x=cy-y\log y \).

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: (d) \( (y-x)^2(-x^2-1+ce^{x^2})=1 \)

Explanation:
1. Put \( v=y-x \): \( v'=-xv-x^3v^3 \).
2. With \( w=v^{-2} \): \( w'-2xw=2x^3 \).
3. IF \( e^{-x^2} \): \( we^{-x^2}=-(x^2+1)e^{-x^2}+c \).
4. So \( \frac{1}{(y-x)^2}=-x^2-1+ce^{x^2} \).

Teacher's Note:
1. A Bernoulli equation dy/dx + Py = Qyⁿ is made linear by dividing by yⁿ and putting z = y to the power 1 - n.
2. Solve the linear equation in z and then go back to y.

Answer: (a) \( y^{-5}x^{-5}=\frac52x^{-2}+c \)

Explanation:
1. Divide by \( xy^6 \) and put \( u=y^{-5} \): \( u'-\frac5xu=-5x^2 \).
2. IF \( x^{-5} \): \( ux^{-5}=-5\int x^{-3}dx=\frac52x^{-2}+c \).
3. So \( y^{-5}x^{-5}=\frac52x^{-2}+c \).

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: (a) \( 2\ln18 \)

Explanation:
1. \( \frac{dp}{p-900}=0.5\,dt \).
2. \( 2\ln|p-900|=t+c \); at t = 0, \( c=2\ln50 \).
3. Put p = 0: \( t=2\ln900-2\ln50=2\ln18 \).

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.