JEE Mathematics Differential Equations MCQs Set 26

Answer: (d) order = 4, degree is not defined

Explanation:
1. The highest derivative is \(\frac{d^4y}{dx^4}\), so the order is 4.
2. The derivative \(\frac{d^3y}{dx^3}\) appears inside \(\sin\), so the equation is not a polynomial in the derivatives.
3. Hence the degree is not defined.

Teacher's Note:
1. Degree is defined only when the equation is a polynomial in the derivatives. Trigonometric functions of derivatives break this.

Answer: (a) order = 2, degree = 1

Explanation:
1. The highest-order derivative is \(\frac{d^2s}{dt^2}\), so the order is 2.
2. Its power is 1, so the degree is 1 (the fourth power is on a lower-order derivative).

Teacher's Note:
1. Degree is the power of the highest-order derivative only.

Answer: (c) order = 2, degree = 1

Explanation:
1. The highest derivative is of order 2, and it appears to the first power.
2. \(\sin y\) involves y, not a derivative, so it does not affect the degree.
3. Order 2, degree 1.

Teacher's Note:
1. Only derivatives matter for degree. A function of y alone is allowed.

Answer: (c) 3

Explanation:
1. Differentiating, \(\frac{dy}{dx}=A\).
2. Eliminating A: \(y=x\frac{dy}{dx}+\left(\frac{dy}{dx}\right)^3\).
3. The highest power of the derivative is 3, so the degree is 3.

Teacher's Note:
1. Eliminate the arbitrary constant by substitution and read off the power of the derivative.

Answer: (b) 4

Explanation:
1. \(c_1e^{2x+c_2}=(c_1e^{c_2})e^{2x}\) has only one independent constant.
2. \(c_4\sin(x+c_5)=(c_4\cos c_5)\sin x+(c_4\sin c_5)\cos x\) has two independent constants.
3. The independent constants are 1 + 1 + 2 = 4, so the order is 4.

Teacher's Note:
1. Count only independent arbitrary constants. Combine constants that appear as products or sums.

Answer: (c) 3

Explanation:
1. \(c_1+c_2\) is a single constant, and \(c_3\) another: two constants in the cosine term.
2. \(c_4e^{c_5}\) is one constant.
3. The total is 3 independent constants, so the order is 3.

Teacher's Note:
1. Merge constants that cannot be separated. The order equals the number of independent constants.

Answer: (b) 2

Explanation:
1. The tangent is \(y=mx+\frac am\) with one parameter m, so the equation is of first order.
2. Since \(m=\frac{dy}{dx}\): \(y\frac{dy}{dx}=x\left(\frac{dy}{dx}\right)^2+a\).
3. The highest power of the derivative is 2, so the degree is 2.

Teacher's Note:
1. A one-parameter family gives a first order equation. Clear fractions before reading off the degree.

Answer: (a) \(y''-2y'+2y=0\)

Explanation:
1. Here \(y=e^x(a\cos x+b\sin x)\) corresponds to roots \(1\pm i\) of the auxiliary equation.
2. The characteristic equation is \((m-1)^2+1=0\), i.e. \(m^2-2m+2=0\).
3. So \(y''-2y'+2y=0\).

Teacher's Note:
1. For \(e^{px}(a\cos qx+b\sin qx)\) the auxiliary equation is \((m-p)^2+q^2=0\).

Answer: (c) \(xy'-2y=0\)

Explanation:
1. The family is \(x^2=4ay\).
2. Differentiating, \(2x=4ay'\), so \(4a=\frac{2x}{y'}\).
3. Substituting: \(x^2=\frac{2x}{y'}y\), i.e. \(xy'-2y=0\).

Teacher's Note:
1. Write the equation with one constant and eliminate it using the first derivative.

Answer: (b) \(\frac{d^2y}{dx^2}-y=0\)

Explanation:
1. \(y'=c_1e^x-c_2e^{-x}\) and \(y''=c_1e^x+c_2e^{-x}=y\).
2. So \(y''-y=0\).

Teacher's Note:
1. Differentiate twice; the second derivative reproduces the function.

Answer: (c) \(\frac{dy}{dx}=\frac{2xy}{x^2-y^2}\)

Explanation:
1. The circles are \(x^2+y^2-2ky=0\).
2. Differentiating, \(2x+2yy'-2ky'=0\), so \(k=\frac{x+yy'}{y'}\).
3. Substituting: \(x^2+y^2=\frac{2y(x+yy')}{y'}\), which gives \((x^2-y^2)y'=2xy\), i.e. \(y'=\frac{2xy}{x^2-y^2}\).

Teacher's Note:
1. Centre on the y-axis and passing through the origin gives \(x^2+y^2-2ky=0\).

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

Explanation:
1. Differentiating once: \(Ax+Byy'=0\). Again: \(A+B\left(y'^2+yy''\right)=0\).
2. Eliminate A and B from these two equations: \(x\left(y'^2+yy''\right)-yy'=0\).
3. That is \(xyy''+xy'^2-yy'=0\).

Teacher's Note:
1. Differentiate twice for two constants. Eliminate by taking the ratio of the two relations.

Answer: (d) \(\frac{dy}{dx}=\frac{1-x}{y-2}\)

Explanation:
1. The circles are \((x-1)^2+(y-2)^2=r^2\).
2. Differentiating, \((x-1)+(y-2)y'=0\).
3. So \(\frac{dy}{dx}=\frac{1-x}{y-2}\).

Teacher's Note:
1. Differentiating removes the radius, the only arbitrary constant.

Answer: (a) \((x^2-y^2)\frac{dy}{dx}=2xy\)

Explanation:
1. \(2a=\frac{x^2+y^2}{y}\). Differentiating \(x^2+y^2-2ay=0\) gives \(2x+2yy'-2ay'=0\), i.e. \(x+yy'=ay'\).
2. Substitute \(a=\frac{x^2+y^2}{2y}\): \(2xy+2y^2y'=(x^2+y^2)y'\).
3. So \((x^2-y^2)y'=2xy\).

Teacher's Note:
1. The family is the circles through the origin with centre on the y-axis; the same equation as an earlier question.

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

Explanation:
1. Differentiating, \(2yy'=4a\), so \(a=\frac{yy'}{2}\).
2. Substituting: \(y^2=2yy'\left(x+\frac{yy'}{2}\right)=2xyy'+y^2y'^2\).
3. Dividing by y: \(2xy'+yy'^2-y=0\).

Teacher's Note:
1. Eliminate a using the first derivative, then simplify.

Answer: (b) \(x\frac{dy}{dx}=-y\)

Explanation:
1. The hyperbola is \(xy=c^2\).
2. Differentiating, \(x\frac{dy}{dx}+y=0\), i.e. \(x\frac{dy}{dx}=-y\).

Teacher's Note:
1. Take the asymptotes as the coordinate axes. The equation then becomes \(xy= ext{constant}\).

Answer: (a) \(y^2=x^2+2xy\frac{dy}{dx}\)

Explanation:
1. The circles are \(x^2+y^2-2ax=0\).
2. Differentiating, \(2x+2yy'-2a=0\), so \(a=x+yy'\).
3. Substituting: \(x^2+y^2=2x(x+yy')\), i.e. \(y^2=x^2+2xyy'\).

Teacher's Note:
1. The centre \((a,0)\) lies on the x-axis, and the circle passes through the origin.

Answer: (b) \(2a\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^3=0\)

Explanation:
1. The family is \((y-k)^2=4a(x-h)\). Differentiating, \(2(y-k)y'=4a\).
2. Differentiating again: \(2y'^2+2(y-k)y''=0\). Using \(y-k=\frac{2a}{y'}\): \(2y'^2+\frac{4ay''}{y'}=0\).
3. Multiplying by \(\frac{y'}{2}\): \(y'^3+2ay''=0\).

Teacher's Note:
1. Two constants (h and k) need two differentiations.

Answer: (c) \(\cos\left(\frac{y-2}{x}\right)=a\)

Explanation:
1. \(\frac{dy}{dx}=\cos^{-1}a\), so \(y=x\cos^{-1}a+C\).
2. At \(x=0\), \(y=2\) gives \(C=2\), so \(\frac{y-2}{x}=\cos^{-1}a\).
3. Hence \(\cos\left(\frac{y-2}{x}\right)=a\).

Teacher's Note:
1. Integrate directly since \(\cos^{-1}a\) is a constant.

Answer: (a) \(2y-1=e^x(\sin x-\cos x)\)

Explanation:
1. \(\int e^x\sin x\,dx=\frac{e^x}{2}(\sin x-\cos x)+C\).
2. At \((0,0)\): \(0=\frac12(0-1)+C\), so \(C=\frac12\).
3. Hence \(2y-1=e^x(\sin x-\cos x)\).

Teacher's Note:
1. Use the standard integral of \(e^x\sin x\), then fix the constant by the point.

Answer: (d) \(\cot\left(\frac yx\right)=\log(ex)\)

Explanation:
1. Put \(y=vx\): \(\frac{dy}{dx}=v-\sin^2v\), so \(v+x\frac{dv}{dx}=v-\sin^2v\).
2. \(\csc^2v\,dv=-\frac{dx}{x}\), giving \(-\cot v=-\log x+c\), i.e. \(\cot\frac yx=\log x+C\).
3. At \(x=1\), \(v=\frac\pi4\): \(1=C\). So \(\cot\frac yx=\log x+1=\log(ex)\).

Teacher's Note:
1. Homogeneous equation: substitute \(y=vx\) and separate the variables.

Answer: (c) \(y=\frac{2x}{1-\log|x|}\)

Explanation:
1. Put \(y=vx\): \(v+x\frac{dv}{dx}=v+\frac{v^2}{2}\), so \(\frac{dv}{v^2}=\frac{dx}{2x}\).
2. \(-\frac1v=\frac12\log|x|+C\). At \(x=1\), \(v=2\): \(C=-\frac12\).
3. \(v=\frac{2}{1-\log|x|}\), so \(y=\frac{2x}{1-\log|x|}\).

Teacher's Note:
1. Homogeneous equation; the constant is fixed using \(v=rac yx=2\) at \(x=1\).

Answer: (b) \(y=4\sin^3x-2\sin^2x\)

Explanation:
1. The integrating factor is \(e^{-3\int\cot x\,dx}=\sin^{-3}x\).
2. \(\frac{d}{dx}\left(\frac{y}{\sin^3x}\right)=\frac{\sin2x}{\sin^3x}=\frac{2\cos x}{\sin^2x}\), so \(\frac{y}{\sin^3x}=-\frac{2}{\sin x}+C\).
3. At \(x=\frac\pi2\), \(y=2\): \(2=-2+C\), \(C=4\). So \(y=4\sin^3x-2\sin^2x\).

Teacher's Note:
1. Linear equation: multiply by the integrating factor and integrate.

Answer: (a) \(\tan^{-1}y+\tan^{-1}e^x=\frac\pi2\)

Explanation:
1. Separate: \(\frac{dy}{1+y^2}=-\frac{e^xdx}{1+e^{2x}}\).
2. Integrating, \(\tan^{-1}y=-\tan^{-1}e^x+C\), i.e. \(\tan^{-1}y+\tan^{-1}e^x=C\).
3. At \((0,1)\): \(\frac\pi4+\frac\pi4=\frac\pi2=C\).

Teacher's Note:
1. The variables separate directly. Use the point to find the constant.

Answer: (a) \(\log|x-y|=x+y+1\)

Explanation:
1. Put \(z=x-y\), so \(dy=dx-dz\). The equation becomes \(z(2dx-dz)=dz\), i.e. \(2z\,dx=(1+z)dz\).
2. \(2dx=\left(1+\frac1z\right)dz\), giving \(2x=z+\log|z|+C\).
3. At \(x=0\), \(z=1\): \(C=-1\). So \(2x=x-y+\log|x-y|-1\), i.e. \(\log|x-y|=x+y+1\).

Teacher's Note:
1. Substituting \(z=x-y\) makes the variables separable.

Answer: (d) \(\tan\left(\frac{y}{2x}\right)=-\frac{1}{2x^2}+\frac32\)

Explanation:
1. Put \(y=vx\): \(x^2(v+xv')-x^2v=1+\cos v\), so \(x^3v'=1+\cos v\).
2. \(\frac{dv}{1+\cos v}=\frac{dx}{x^3}\), i.e. \(\frac12\sec^2\frac v2dv=\frac{dx}{x^3}\). Integrating, \(\tan\frac v2=-\frac{1}{2x^2}+C\).
3. At \(x=1\), \(v=\frac\pi2\): \(1=-\frac12+C\), so \(C=\frac32\). Hence \(\tan\frac{y}{2x}=-\frac{1}{2x^2}+\frac32\).

Teacher's Note:
1. Use \(1+\cos v=2\cos^2rac v2\) to integrate the left side.

Answer: (a) \(\frac13\)

Explanation:
1. Separate: \(\frac{dy}{1+y}=-\frac{\cos x}{2+\sin x}dx\), so \(\log(1+y)=-\log(2+\sin x)+c\).
2. \((1+y)(2+\sin x)=C\). At \(x=0\), \(y=1\): \(C=4\).
3. At \(x=\frac\pi2\): \((1+y)\cdot3=4\), so \(y=\frac13\).

Teacher's Note:
1. Combine logarithms into a product, then use the initial condition.

Answer: (c) \(-\frac12\)

Explanation:
1. Write as \(\frac{dy}{dt}-\frac{t}{1+t}y=\frac{1}{1+t}\). The integrating factor is \(e^{-\int\frac{t}{1+t}dt}=(1+t)e^{-t}\).
2. \(\frac{d}{dt}\left[y(1+t)e^{-t}\right]=e^{-t}\), so \(y(1+t)e^{-t}=-e^{-t}+C\). With \(y(0)=-1\): \(C=0\).
3. So \(y=-\frac{1}{1+t}\) and \(y(1)=-\frac12\).

Teacher's Note:
1. Compute the integrating factor by writing \(rac{t}{1+t}=1-rac{1}{1+t}\).

Answer: (b) a parabola whose axis is parallel to the y-axis

Explanation:
1. Integrating twice, \(y=x^2+c_1x+c_2\).
2. This is a parabola with its axis parallel to the y-axis.

Teacher's Note:
1. A quadratic polynomial in x represents a parabola opening up or down.

Answer: (d) \(y=-\frac12(\log x)^2-\log x\)

Explanation:
1. \(y''=\frac{\log x}{x^2}\), so \(y'=-\frac{\log x+1}{x}+C\). At \(x=1\), \(y'=-1\) gives \(C=0\).
2. \(y=-\int\frac{\log x}{x}dx-\int\frac{dx}{x}=-\frac12(\log x)^2-\log x+D\), and \(y(1)=0\) gives \(D=0\).
3. So \(y=-\frac12(\log x)^2-\log x\).

Teacher's Note:
1. Integrate twice and fix each constant using the given conditions.