JEE Mathematics Differential Equations MCQs Set 21

Answer: (a) \( 6x\sec y=x^6+C \)

Explanation:
1. Divide by \( \cos^2y \): \( \sec y\tan y\,y'+\frac1x\sec y=x^4 \).
2. Put \( w=\sec y \): \( w'+\frac wx=x^4 \).
3. Multiply by x: \( (xw)'=x^5 \), so \( xw=\frac{x^6}{6}+c \), i.e. \( 6x\sec y=x^6+C \).

Teacher's Note:
1. Divide through by cos^2 y first.
2. Then w = sec y gives a linear equation.

Answer: (b) \( x^2y+x^3=\frac yx+C \)

Explanation:
1. Write \( (2x^3y+3x^4+y)dx-(x-x^4)dy=0 \).
2. Divide by \( x^2 \): \( \left(2xy+3x^2+\frac y{x^2}\right)dx+\left(x^2-\frac1x\right)dy=0 \), which is exact.
3. It equals \( d\left(x^2y+x^3-\frac yx\right)=0 \).

Teacher's Note:
1. Dividing by x^2 makes the equation exact.
2. Check by differentiating the option.

Answer: (a) \( \ln(xy)=\sin^{-1}x+C \)

Explanation:
1. Rewrite: \( x\,dy+y\,dx=\frac{xy}{\sqrt{1-x^2}}dx \).
2. Divide by xy: \( \frac{d(xy)}{xy}=\frac{dx}{\sqrt{1-x^2}} \).
3. Integrate: \( \ln(xy)=\sin^{-1}x+C \).

Teacher's Note:
1. x dy + y dx is d(xy).
2. Divide by xy to separate.

Answer: (b) \( x^3y=\frac{y}{x^2}+C \)

Explanation:
1. Left side: \( x^3(x^3dy+3x^2y\,dx)=x^3\,d(x^3y) \).
2. Right side: \( x\,dy-2y\,dx=x^3\,d\left(\frac y{x^2}\right) \).
3. Cancel \( x^3 \): \( d(x^3y)=d\left(\frac y{x^2}\right) \), so \( x^3y=\frac y{x^2}+C \).

Teacher's Note:
1. Look for exact differentials on both sides.
2. Both sides carry a common factor x^3.

Answer: (d) \( y=3x+9 \)

Explanation:
1. The equation is \( y=xp+p^2 \) with \( p=y' \), a Clairaut equation (rewrite \( p^2+xp-y=0 \) as \( y=xp+p^2 \)).
2. Its general solution is \( y=cx+c^2 \).
3. For c = 3: \( y=3x+9 \); the other options do not fit \( y=cx+c^2 \).

Teacher's Note:
1. The equation has the form y = x p + f(p).
2. Replace p by a constant c.

Answer: (b) \( y=x^3+3x+1 \)

Explanation:
1. Separate: \( \frac{y_2}{y_1}=\frac{2x}{x^2+1} \), so \( y_1=a(x^2+1) \).
2. At x = 0 the slope is 3: a = 3, so \( y_1=3x^2+3 \).
3. Integrate: \( y=x^3+3x+b \); y(0) = 1 gives b = 1.

Teacher's Note:
1. Treat y' as the unknown and integrate twice.
2. Use both given conditions.

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

Explanation:
1. These parabolas are \( y^2=4a(x-h) \), with two constants.
2. Differentiate: \( 2yy'=4a \) and again \( y'^2+yy''=0 \).
3. This eliminates a and h.

Teacher's Note:
1. Two constants require two differentiations.
2. The second derivative of y^2 is 2(y'^2 + y y'').

Answer: (c) \( \frac{dy}{dx}+f(x)y=2r(x) \)

Explanation:
1. Add the two identities: \( y_1'+fy_1=r \) and \( y_2'+fy_2=r \).
2. The sum is \( (y_1+y_2)'+f(y_1+y_2)=2r \).

Teacher's Note:
1. The operator on the left is linear.
2. Add the two equations.

Answer: (c) \( y_1\int\frac{r(x)}{y_1}dx \)

Explanation:
1. Try \( y=y_1u \): \( y'+fy=y_1'u+y_1u'+fy_1u=y_1u' \) since \( y_1'+fy_1=0 \).
2. So \( y_1u'=r \) and \( u=\int\frac r{y_1}dx \).
3. Hence \( y=y_1\int\frac{r}{y_1}dx \).

Teacher's Note:
1. This is variation of parameters.
2. Substitute y = y1 u into the equation.

Answer: (d) \( \frac xy+e^{x^3}=C \)

Explanation:
1. Divide by \( y^2 \): \( \frac{y\,dx-x\,dy}{y^2}+3x^2e^{x^3}dx=0 \).
2. The first term is \( d\left(\frac xy\right) \), the second is \( d(e^{x^3}) \).
3. So \( \frac xy+e^{x^3}=C \).

Teacher's Note:
1. Divide by y^2 to recognise d(x/y).
2. The second term is an exact derivative of e^(x^3).

Answer: (d) \( x^3\sin^3y=3y^2\sin x+C \)

Explanation:
1. Check exactness: \( \frac{\partial M}{\partial y}=3x^2\sin^2y\cos y-2y\cos x=\frac{\partial N}{\partial x} \).
2. Integrate M with respect to x: \( \frac{x^3\sin^3y}{3}-y^2\sin x \).
3. So \( x^3\sin^3y-3y^2\sin x=C \).

Teacher's Note:
1. Test M_y = N_x.
2. Integrate M with respect to x, treating y as constant.

Answer: (d) \( x^{m+1}y^{n+1}+1=cxy \)

Explanation:
1. Use \( d(x^my^n)=x^{m-1}y^{n-1}(my\,dx+nx\,dy) \).
2. So the left side times \( x^my^n \) is \( x^2y^2\,d(x^my^n) \), and the equation becomes \( x^2y^2\,du=d(xy) \) with \( u=x^my^n \).
3. Hence \( du=\frac{d(xy)}{(xy)^2} \), \( u=-\frac1{xy}+c \), i.e. \( x^{m+1}y^{n+1}+1=cxy \).

Teacher's Note:
1. Use u = x^m y^n and the product d(xy).
2. Multiply the equation by x^m y^n first.

Answer: (a) \( x=y(b-a\ln y) \)

Explanation:
1. The tangent at (x, y) meets the x-axis at \( x-\frac{y}{y'} \), so \( x-y\frac{dx}{dy}=ay \).
2. Then Rearranged, \( y\frac{dx}{dy}-x=-ay \), so \( \frac{d}{dy}\left(\frac xy\right)=\frac{y\,x'-x}{y^2}=-\frac{a}{y} \).
3. Integrating, \( \frac xy=b-a\ln y \), so \( x=y(b-a\ln y) \).

Teacher's Note:
1. Find the x-intercept of the tangent.
2. Treat x as the dependent variable.

Answer: (b) \( x^2+y^2=2x \)

Explanation:
1. The normal at (x, y) is \( (X-x)+y'(Y-y)=0 \); its distance from the origin is \( \frac{|x+yy'|}{\sqrt{1+y'^2}}=|y| \).
2. Squaring: \( x^2+2xyy'=y^2 \), a homogeneous equation with solution \( x^2+y^2=cx \).
3. Through (1, 1): c = 2, so \( x^2+y^2=2x \).

Teacher's Note:
1. Write the equation of the normal.
2. The resulting equation is homogeneous.

Answer: (b) \( f(x)\ge e^{3x}\ \text{for all }x\ge0 \)

Explanation:
1. Let \( g(x)=e^{-3x}f(x) \). Then \( g'(x)=e^{-3x}(f'-3f)>0 \).
2. So g is increasing: \( g(x)\ge g(0)=1 \).
3. Hence \( f(x)\ge e^{3x} \) (equality only at x = 0, so strict inequality fails at 0).

Teacher's Note:
1. Consider g(x) = f(x) e^(-3x).
2. Check x = 0 separately for strict inequality.

Answer: (a) \( y=\frac{c}{x^3}e^{-1/x} \)

Explanation:
1. Differentiate: \( \int_0^xy\,dt+xy=\int_0^xty\,dt+(x+1)xy \), so \( \int_0^xy\,dt=\int_0^xty\,dt+x^2y \).
2. Differentiate again: \( y=xy+x^2y'+2xy \), i.e. \( \frac{y'}{y}=\frac1{x^2}-\frac3x \).
3. Integrate: \( \ln y=-\frac1x-3\ln x+c \), so \( y=\frac{c}{x^3}e^{-1/x} \).

Teacher's Note:
1. Differentiate the integral equation twice using the Leibniz rule.
2. The result is a separable equation.

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

Explanation:
1. Write \( y'+\frac{x}{1-x^2}y=\frac{ax}{1-x^2} \); the integrating factor is \( (1-x^2)^{-1/2} \).
2. Then \( \frac{d}{dx}\left[\frac{y}{\sqrt{1-x^2}}\right]=\frac{ax}{(1-x^2)^{3/2}} \), so \( \frac{y}{\sqrt{1-x^2}}=\frac{a}{\sqrt{1-x^2}}+c \).
3. So \( y-a=c\sqrt{1-x^2} \) and \( (y-a)^2+c^2x^2=c^2 \).

Teacher's Note:
1. Find the integrating factor exp of the integral of x/(1-x^2).
2. Square the final relation to remove the root.

Answer: (d) \( \frac{27}{8} \)

Explanation:
1. Differentiate: \( 2c(y+c)y'=3x^2 \); divide by the curve equation: \( \frac{2y'}{y+c}=\frac3x \), so \( c=\frac{2xy'}{3}-y \).
2. Substitute in \( c(y+c)^2=x^3 \): \( \left(\frac{2xy'}3-y\right)\frac{4x^2y'^2}{9}=x^3 \).
3. Simplify: \( 12y\,y'^2+27x=8x\,y'^3 \), so a = 27, b = 8 and a/b = 27/8.

Teacher's Note:
1. Eliminate c using the derivative.
2. Compare coefficients with the given form.

Answer: (a) 2

Explanation:
1. Treat x as dependent: \( \frac{dx}{dy}-x\cos y=2\sin y\cos y \).
2. Integrating factor \( e^{-\sin y} \): \( xe^{-\sin y}=\int2\sin y\cos y\,e^{-\sin y}dy \).
3. With \( u=\sin y \): \( \int2ue^{-u}du=-2(u+1)e^{-u}+c \).
4. So \( x=ce^{\sin y}-2(1+\sin y) \) and k = 2.

Teacher's Note:
1. Invert the derivative to get a linear equation in x.
2. Integrate by parts after u = sin y.

Answer: (a) 1

Explanation:
1. Substituting the combination: \( \alpha(y_1'+Py_1)+\beta(y_2'+Py_2)=(\alpha+\beta)Q \).
2. This must equal Q, so \( \alpha+\beta=1 \).

Teacher's Note:
1. Use the fact that y1 and y2 each satisfy the equation.
2. Q is not identically zero.

Answer: (c) 16

Explanation:
1. Factorise: \( (y y'+x)(y'-1)=0 \).
2. The factor \( yy'+x=0 \) gives circles \( x^2+y^2=c \), not lines, so the line comes from \( y'=1 \): \( y=x+k \).
3. Through (3, 4): \( y=x+1 \), i.e. \( 2x-2y+2=0 \), so A = 2, B = -2 and \( A^3-B^3=8+8=16 \).

Teacher's Note:
1. Factor the quadratic in y'.
2. Only y' = 1 gives a line.

Answer: (a) 25

Explanation:
1. The distance condition is \( \frac{|y-xy'|}{\sqrt{1+y'^2}}=|x| \), giving \( y^2-2xyy'=x^2 \), with solution \( x^2+y^2=cx \).
2. The curve passes through (1, 3): c = 10. Slope there: \( 2+6y'=10 \), so \( y'=\frac43 \).
3. The tangent is \( 4x-3y+5=0 \), so a = 4, b = -3 and \( a^2+b^2=25 \).

Teacher's Note:
1. Use the point-to-line distance formula to form the ODE.
2. The ODE is homogeneous.