JEE Mathematics Differential Equations MCQs Set 16

Answer: (b) \( \left(x^2+1+ce^{x^2}\right)y^2=1 \)

Explanation:
1. Divide by \( y^3 \) and put \( u=y^{-2} \): \( u'-2xu=-2x^3 \).
2. IF \( e^{-x^2} \): \( ue^{-x^2}=\int-2x^3e^{-x^2}dx=(x^2+1)e^{-x^2}+c \).
3. \( \frac1{y^2}=x^2+1+ce^{x^2} \).

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: (b) \( g(x)+\log(1+y-g(x))=c \)

Explanation:
1. IF \( =e^{g} \).
2. \( ye^g=\int gg'e^gdx=(g-1)e^g+c \).
3. \( y-g+1=ce^{-g} \).
4. Taking logs: \( g+\log(1+y-g)=\log c \).

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) \( a\frac{d^2y}{dx^2}+x^2\frac{dy}{dx}+y=0 \)

Explanation:
1. Substitute f and g and subtract.
2. The right-hand sides cancel: \( e^x-e^x=0 \).
3. So \( f-g \) satisfies the homogeneous equation \( ay''+x^2y'+y=0 \).

Teacher's Note:
1. Recall the standard result for the type of equation and match it with the given form.
2. Eliminating the constants or integrating directly confirms it.

Answer: (c) \( y(x+c)=f(x) \)

Explanation:
1. Write \( f\,dy=(yf'-y^2)dx \).
2. Divide by \( y^2 \): \( \frac{f\,dy-yf'dx}{y^2}=-dx \).
3. The left side is \( -d\!\left(\frac fy\right) \).
4. \( \frac fy=x+c \), so \( y(x+c)=f(x) \).

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

Answer: (c) 1

Explanation:
1. Put \( x=\tan a,\ y=\tan b \): \( \cos a+\cos b=K(\sin a-\sin b) \).
2. This gives \( \cot\frac{a-b}{2}=K \), so \( \tan^{-1}x-\tan^{-1}y=\text{constant} \).
3. Differentiating: \( \frac{dx}{1+x^2}=\frac{dy}{1+y^2} \) which is of order 1 and degree 1.

Teacher's Note:
1. The order is the highest derivative present; the degree is its power after the equation is made a polynomial in the derivatives.
2. Remove radicals and fractional powers before reading off the degree.

Answer: (b) \( \log\left|\tan\left(\frac{x^2+y^2}{2}\right)\right|=\frac{x^2}{y^2}+c \)

Explanation:
1. Use \( x\,dx+y\,dy=\tfrac12d(x^2+y^2) \) and \( y\,dx-x\,dy=y^2d\!\left(\frac xy\right) \).
2. \( \frac{d(x^2+y^2)}{2}=\frac xy\sin(x^2+y^2)\,d\!\left(\frac xy\right) \).
3. \( \frac{dt}{\sin t}=2\frac xy\,d\!\left(\frac xy\right)=d\!\left(\frac{x^2}{y^2}\right) \) with \( t=x^2+y^2 \).
4. \( \log\left|\tan\frac t2\right|=\frac{x^2}{y^2}+c \).

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: (d) a hyperbola

Explanation:
1. The tangent meets the x-axis at \( x-\frac{y}{y_1} \).
2. Bisection means this equals 2x, so \( -\frac{y}{y_1}=x \).
3. \( x\,y_1+y=0 \), so \( xy=c \), 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: (b) \( y\sin x+x\sin y+xy=k \)

Explanation:
1. Write \( (y\cos x+\sin y+y)dx+(\sin x+x\cos y+x)dy=0 \).
2. \( M_y=\cos x+\cos y+1=N_x \), so it is exact.
3. \( F=y\sin x+x\sin y+xy \).

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: (d) a-4, b-1, c-2, d-3

Explanation:
1. (a) Slope 2 means \( \frac{dy}{dx}=2 \), which is (4).
2. (b) \( y=ax^2+bx+c \) gives \( y_3=0 \), which is (1).
3. (c) \( xy=c^2 \) gives \( x\,dy+y\,dx=0 \), which is (2).
4. (d) \( \log y=cx \) gives \( \frac{y_1}{y}=c=\frac{\log y}{x} \), i.e. \( xy_1=y\log y \), which is (3).

Teacher's Note:
1. Work out each item of List-I on its own and then look for the pairing among the options.
2. Eliminating options with one certain pairing often settles the question quickly.

Answer: (c) a-4, b-5, c-1, d-3

Explanation:
1. (a) \( y\,dy=\sec^2x\,dx \) gives \( y^2=2\tan x+c \), which is (4).
2. (b) \( \cos y\,dy=x\,dx \) gives \( x^2=2\sin y+c \), which is (5).
3. (c) IF \( \sec^2x \): \( y\sec^2x=\int\sin x\sec^2x\,dx=\sec x+c \), which is (1).
4. (d) \( \frac{d(xy)}{dx}=\cos x \) gives \( xy=\sin x+c \), which is (3).

Teacher's Note:
1. Solve or classify every entry before matching.
2. Check the final pairing against all four rows.

Answer: (a) only I

Explanation:
1. I: \( \frac{dy}{y}=\frac{x\,dx}{1+x^2} \) gives \( \log y=\frac12\log(1+x^2)+\log c \), so I is true.
2. II: \( y\,dy=(1-x)dx \) gives \( y^2=2x-x^2+c \) and not \( y^2=(1-x)^2 \), so II is false.

Teacher's Note:
1. Test each statement separately by solving or differentiating.
2. The answer depends on which statements survive the check.

Answer: (d) A-3, B-4, C-2, D-1

Explanation:
1. A: \( y_2=y \), i.e. (3).
2. B: \( y_2=-p^2y \), i.e. (4).
3. C: \( y_3=0 \), i.e. (2).
4. D: \( y_2+m^2y=0 \), i.e. (1).

Teacher's Note:
1. Solve or classify every entry before matching.
2. Check the final pairing against all four rows.

Answer: (a) only I

Explanation:
1. I: \( \frac{dx}{dy}+x\tan y=\sec y \), IF \( =e^{\int\tan y\,dy}=\sec y \), true.
2. II: \( \frac{dy}{dx}-\frac{y}{x(x-1)}=x(x-1) \) and \( \int\frac{-dx}{x(x-1)}=-\log\frac{x-1}{x} \).
3. So IF \( =\frac{x}{x-1} \), not \( \frac{x-1}{x} \); II is false.

Teacher's Note:
1. Test each statement separately by solving or differentiating.
2. The answer depends on which statements survive the check.

Answer: (c) only A is true

Explanation:
1. The correct differential is \( d\!\left(\tan^{-1}\frac yx\right)=\frac{x\,dy-y\,dx}{x^2+y^2} \), so R is false (its denominator is xy).
2. Hence \( d\left(\frac{x^2+y^2}{2}\right)=d\left(\tan^{-1}\frac yx\right) \), so \( x^2+y^2=2\tan^{-1}\frac yx+c \).
3. A is true.

Teacher's Note:
1. Verify the statement by an independent calculation rather than by recalling a formula.
2. A single wrong sign makes a statement false.

Answer: (b) A-3, B-5, C-1, D-2

Explanation:
1. A: \( e^{-\int\cot x}=\mathrm{cosec}\,x \), (3).
2. B: \( e^{\int\frac{dx}{x\log x}}=\log x \), (5).
3. C: \( P=\cot x+\frac1x \), IF \( x\sin x \), (1).
4. D: \( P=\frac1x+1 \), IF \( xe^x \), (2).

Teacher's Note:
1. Work out each item of List-I on its own and then look for the pairing among the options.
2. Eliminating options with one certain pairing often settles the question quickly.

Answer: (c) both I and II

Explanation:
1. I: \( \frac{dx}{dy}-\frac xy=2y^2 \), IF \( \frac1y \): \( \frac xy=y^2+c \), so I is true.
2. II: IF \( \sec x \): \( y\sec x=\int\cos x\,dx=\sin x+c \), so II is true.

Teacher's Note:
1. Verify the statement by an independent calculation rather than by recalling a formula.
2. A single wrong sign makes a statement false.

Answer: (a) A and R are both true and R is the correct explanation of A

Explanation:
1. IF \( =(x+1)^{-2} \).
2. \( \frac{y}{(x+1)^2}=\int(x+1)dx=\frac{(x+1)^2}{2}+c \).
3. Multiply by 2: \( \frac{2y}{(x+1)^2}=(x+1)^2+c' \), so A is true, and it follows from the rule in R.

Teacher's Note:
1. Test each statement separately by solving or differentiating.
2. The answer depends on which statements survive the check.

Answer: (c) A true, R true

Explanation:
1. Here p = 1, so \( e^{\int1\,dx}=e^x \).
2. A is true.
3. R states the standard formula, so it is also true.

Teacher's Note:
1. Verify the statement by an independent calculation rather than by recalling a formula.
2. A single wrong sign makes a statement false.

Answer: (c) I is true and II is false

Explanation:
1. I: multiply by \( e^{x^2} \): \( d(ye^{x^2})=2\,dx \), so \( ye^{x^2}=2x+c \); true.
2. II: differentiating \( ye^{x^2}+2x=c \) gives \( e^{x^2}dy+2xye^{x^2}dx+2\,dx=0 \), which does not give the stated relation; false.

Teacher's Note:
1. Test each statement separately by solving or differentiating.
2. The answer depends on which statements survive the check.

Answer: (d) \( yy''=(y')^2 \)

Explanation:
1. \( \log y=\log c_1+c_2x \).
2. Differentiate: \( \frac{y'}{y}=c_2 \).
3. Differentiate again: \( \frac{yy''-y'^2}{y^2}=0 \).
4. So \( yy''=(y')^2 \).

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: (d) 7

Explanation:
1. \( \frac{dy}{y+3}=dx \).
2. \( \ln(y+3)=x+c \); at x = 0: \( y+3=5e^x \).
3. At \( x=\ln2 \): \( y+3=10 \), so y = 7.

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

Answer: (c) 3500

Explanation:
1. \( P=2000+\int_0^{25}(100-12\sqrt x)dx \).
2. \( =2000+\left[100x-8x^{3/2}\right]_0^{25} \).
3. \( =2000+2500-1000=3500 \).

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.