Chapter 17
A Complete Resource Book in Mathematics for JEE Main 2017 · 93 exercises
Problem 62
The solution of the equation \(\frac{d y}{d x}=\frac{y}{2 y \log y+y-x}\) (A) \(x=y \log y+\frac{c}{y}\) (B) \(y=x \log x+\frac{c}{x}\) (C) \(x=-y \log y+\frac{c}{y}\) (D) None of these
7 step solution
Problem 63
Solution of the equation \(\cos ^{2} x \frac{d y}{d x}-y \tan 2 x=\cos ^{4} x\), when \(|x|<\frac{\pi}{4}\) and \(y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}\), (A) \(y=\frac{\sin 2 x}{2\left(\tan ^{2} x-1\right)}\) (B) \(y=\frac{\sin 2 x}{2\left(1-\tan ^{2} x\right)}\) (C) \(y=\frac{\sin 2 x}{2\left(1+\tan ^{2} x\right)}\) (D) None of these
9 step solution
Problem 64
The solution of the equation \(\frac{d y}{d x}+x(x+y)=\) \(x^{3}(x+y)^{3}-1\) is (A) \((x+y)^{-3}=c e^{x^{2}}+x^{2}+1\) (B) \((x+y)^{-2}=c e^{x 2}-x^{2}+1\) (C) \((x+y)^{-2}=c e^{x 2}+x^{2}+1\) (D) None of these
7 step solution
Problem 65
The solution of the equation \(\sin y \frac{d y}{d x}=\cos y(1-x\) \(\cos y\) ) is (A) \(\sec y=(1+x)+c e^{x}\) (B) \(\tan y=(1+x)+c e^{x}\) (C) \(\sec y=(1+x)+c e^{-x}\) (D) None of these
5 step solution
Problem 66
The solution of the differential equation \(x\left(y^{2} e^{x y}+e^{x y}\right) d y=y\left(e^{x / y}-y^{2} e^{x y}\right) d x\) is (A) \(x y=\ln \left(e^{y / x}+c\right)\) (B) \(x y=\ln \left(e^{x / y}+c\right)\) (C) \(\frac{y}{x}=\ln \left(e^{x y}+c\right)\) (D) \(\frac{x}{y}=\ln \left(e^{x y}+c\right)\)
6 step solution
Problem 67
Solution of the equation \(x \int_{0}^{x} y(t) d t=(x+1) \int_{0}^{x} t y(t) d t, x>0\) is (A) \(y=\frac{c}{x^{3}} e^{-\frac{1}{x}}\) (B) \(y=\frac{c}{x^{3}} e^{\frac{1}{x}}\) (C) \(y=\frac{c}{x} e^{-\frac{1}{x^{\prime}}}\) (D) \(y=\frac{c}{x} e^{\frac{1}{x^{1}}}\)
5 step solution
Problem 68
The solution of the differential equation \((1+\tan y)(d x-d y)+2 x d y=0\) is (A) \(x(\sin y+\cos y)=\sin y+c e^{-y}\) (B) \(x(\sin y-\cos y)=\sin y+c e^{-y}\) (C) \(x(\sin y+\cos y)=\cos y+c e^{-y}\) (D) None of these
8 step solution
Problem 70
Solution of the differential equation \(2 y \sin x \frac{d y}{d x}=\) \(2 \sin x \cos x-y^{2} \cos x\) satisfying \(y\left(\frac{\pi}{2}\right)=1\) is given by (A) \(y^{2}=\sin x\) (B) \(y=\sin ^{2} x\) (C) \(y^{2}=\cos x+1\) (D) \(y^{2} \sin x=4 \cos ^{2} x\)
8 step solution
Problem 71
The solution of the equation \(y\left(2 x^{2} y+e^{x}\right) d x-\left(e^{x}+y^{3}\right) d y=0\), if \(y(0)=1\), is (A) \(6 e^{x}-4 x^{3} y-3 y^{3}-3 y=0\) (B) \(6 e^{x}+4 x^{3} y-3 y^{3}-3 y=0\) (C) \(6 e^{x}+4 x^{3} y+3 y^{3}-3 y=0\) (D) None of these
5 step solution
Problem 74
If \(y=c_{1} e^{2 x}+c_{2} e^{x}+c_{3} e^{-x}\) satisfies the differential equation \(\frac{d^{3} y}{d x^{3}}+a \frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0\), then \(\frac{a^{3}+b^{3}+c^{3}}{a b c}\) is equal to (A) \(\frac{1}{4}\) (B) \(-\frac{1}{4}\) (C) \(\frac{1}{2}\) (D) \(-\frac{1}{2}\)
7 step solution
Problem 76
If \(g(x)\) be a function defined on \([-1,1]\). If the area of the equilateral triangle with two of its vertices at \((0,0)\) and \((x, g(x))\) is \(\frac{\sqrt{3}}{4}\), then the function is (A) \(g(x)=\pm \sqrt{\left(1-x^{2}\right)}\) (B) \(g(x)=-\sqrt{\left(1-x^{2}\right)}\) (C) \(g(x)=\sqrt{\left(1-x^{2}\right)}\) (D) \(g(x)=\sqrt{\left(1+x^{2}\right)}\)
6 step solution
Problem 77
For a certain curve \(y=f(x)\) satisfying \(\frac{d^{2} y}{d x^{2}}=6 x-4\), \(f(x)\) has a local minimum value 5 when \(x=1\). (A) Equation of the curve is \(y=x^{3}-2 x^{2}+x+5\) (B) \(f(x)\) has a local maximum at \(x=\frac{1}{3}\) (C) Global maximum value of \(f(x)\) is 7 (D) Global minimum value of \(f(x)\) is 5
7 step solution
Problem 78
The solution of the equation \(\frac{d y}{d x}+x=x e^{(n-1) y}\) is (A) \(\frac{1}{(n-1)} \log \left(\frac{e^{(n-1) y}-1}{e^{(n-1) y}}\right)=\frac{x^{2}}{2}+c\) (B) \(e^{(n-1) y}=c e^{(n-1) y+(n-1)^{\frac{x^{2}}{2}}}+1\) (C) \(\log \left(\frac{e^{(n-1) y}-1}{(n-1) e^{(n-1) y}}\right)=n^{2}+c\) (D) \(e^{(n-1) y}=c e^{(n-1) \frac{x^{x}}{2}+x}+1\)
6 step solution
Problem 80
The solution of the equation \(\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x+[x+\log x-x \sin y] d y=0\) is (A) \(y(x+\log x)-x \cos y=c\) (B) \(y(x+\log x)+x \sin y=c\) (C) \(y(x+\log x)+x \cos y=c\) (D) None of these
4 step solution
Problem 81
The solution of the differential equation \(\left(x^{2} y-2 x y^{2}\right) d x-\left(x^{3}-3 x^{2} y\right) d y=0\) is (A) \(\frac{x}{y}-2 \log x+3 \log y=c\) (B) \(\frac{x}{y}+2 \log x+3 \log y=c\) (C) \(\frac{x}{y}-2 \log x-3 \log y=c\) (D) None of these
5 step solution
Problem 82
The solution of the equation \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\) is (A) \(y \sec x y=c x\) (B) \(x \sec x y=c y\) (C) \(x \operatorname{cosec} x y=c y\) (D) None of these
4 step solution
Problem 83
The integrating factor to make the differential equation \(\left(x y^{2}-e^{\frac{1}{x^{2}}}\right) d x-x^{2} y d y=0\) exact is (A) \(\frac{1}{x}\) (B) \(\frac{1}{x^{2}}\) (C) \(\frac{1}{x^{3}}\) (D) \(\frac{1}{x^{4}}\)
6 step solution
Problem 84
The integrating factor to make the differential equation \(\left(y^{4}+2 y\right) d x+\left(x y^{3}+2 y^{4}-4 x\right) d y=0\) exact is (A) \(\frac{1}{y}\) (B) \(\frac{1}{y^{2}}\) (C) \(\frac{1}{y^{3}}\) (D) \(\frac{1}{y^{4}}\)
5 step solution
Problem 87
The solution of the equation \((x-a)\left(\frac{d y}{d x}\right)^{2}+(x-y) \frac{d y}{d x}-y=0\) is (A) \(y=c x+\frac{a c^{2}}{c+1}\) (B) \(y=c x-\frac{a c^{2}}{c+1}\) (C) \(y=c x-\frac{a^{2} c^{2}}{c+1}\) (D) None of these
6 step solution
Problem 88
The solution of the equation \(p^{2} x(x-2)+p(2 y-2 x y-x+2)+y^{2}+y=0\) is (A) \((y+c x+2 c)(y-c x+1)=0\) (B) \((y-c x+2 c)(y+c x+1)=0\) (C) \((y-c x+2 c)(y-c x+1)=0\) (D) \((y-c x+2 c)(y-c x-1)=0\)
5 step solution
Problem 90
The orthogonal trajectory of the family of parabolas \(y^{2}=4 a x\) is (A) \(2 x^{2}+y^{2}=c\) (B) \(x^{2}+2 y^{2}=c\) (C) \(2 x^{2}-y^{2}=c\) (D) None of these
5 step solution
Problem 94
Assertion: The order of the differential equation, of which \(x y=c e^{x}+b e^{-x}+x^{2}\) is a solution, is 2 . Reason: The differential equation is \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0\)
6 step solution
Problem 95
Assertion: A normal is drawn at a point \(P(x, y)\) of a curve. It meets the \(x\)-axis and the \(y\)-axis in points \(A\) and \(B\), respectively, such that \(\frac{1}{O A}+\frac{1}{O B}=1\), where \(O\) is the origin. The equation of such a curve passing through \((5,4)\) is \((x-1)^{2}+(y-1)^{2}=25\). Reason: \(O A=x+y \frac{d y}{d x}\) and \(O B=\frac{\left(x+y \frac{d y}{d x}\right)}{\frac{d y}{d x}}\)
8 step solution
Problem 97
Assertion: The differential equation of all straight lines which are at a constant distance \(p\) from the origin is \(\left(y-x y_{1}\right)^{2}=p^{2}\left(1+y_{1}^{2}\right)\) Reason: The general equation of any straight line which is at a constant distance \(p\) from the origin is \(x \cos \alpha+y \sin \alpha=p .\)
5 step solution
Problem 98
The order and degree of the differential equation \(\left(1+3 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^{3} y}{d x^{3}}\) are \(\quad\) (A) \(\left(1, \frac{2}{3}\right)\) (B) \((3,1)\) (C) \((3,3)\) (D) \((1,2)\)
3 step solution
Problem 99
The solution of the equation \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) is (A) \(\frac{e^{-2 x}}{4}\) (B) \(\frac{e^{-2 x}}{4}+c x+d\) (C) \(\frac{1}{4} e^{-2 x}+c x^{2}+d\) (D) \(\frac{1}{4} e^{-2 x}+c+d\)
4 step solution
Problem 100
The differential equation of all non-vertical lines in a plane is (A) \(\frac{d^{2} y}{d x^{2}}=0\) (B) \(\frac{d^{2} x}{d y^{2}}=0\) (C) \(\frac{d y}{d x}=0\) (D) \(\frac{d x}{d y}=0\)
4 step solution
Problem 101
The degree and order of the differential equation of the family of all parabolas whose axis is \(x\)-axis, are respectively \(\quad\) [2003] (A) 2,1 \(\begin{array}{ll}\text { (B) } 1,2 & \text { (C) } 3,2\end{array}\) (D) 2,3
5 step solution
Problem 102
The solution of the differential equation \(\left(1+y^{2}\right)\left(x-e^{2 \tan ^{-1} y}\right) \frac{d y}{d x}=0\), is \(\quad\) (A) \((x-2)=k e^{-\tan ^{-1} y}\) (B) \(2 x e^{\tan ^{-1}} y=e^{2 \tan ^{-1}}+k\) (C) \(x e^{\operatorname{lan}^{-1} y}=\tan ^{-1} y+k\) (D) \(x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+k\)
4 step solution
Problem 103
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant is (A) \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\) (B) \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\) (C) \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\) (D) \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
6 step solution
Problem 104
The solution of the differential equation \(y d x+(x+\) \(\left.x^{2 y}\right) d y=0\) is (A) \(-\frac{1}{x y}=C\) (B) \(-\frac{1}{x y}+\log y=C\) (C) \(\frac{1}{x y}+\log y=C\) (D) \(\log y=C x\)
8 step solution
Problem 105
The differential equation representing the family of curves \(y^{2}=2 c(x+\sqrt{c})\) where \(c>0\), is a parameter, is of order and degree as follows: \(\quad\) (A) order 1 , degree 2 (B) order 1 , degree 1 (C) order 1 , degree 3 (D) order 2, degree 2
5 step solution
Problem 106
If \(x \frac{d y}{d x}=y(\log y-\log x+1)\), then the solution of the equation is (A) \(y \log \left(\frac{x}{y}\right)=c x\) (B) \(x \log \left(\frac{y}{x}\right)=c y\) (C) \(\log \left(\frac{y}{x}\right)=c x\) (D) \(\log \left(\frac{x}{y}\right)=c y\)
6 step solution
Problem 108
The differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is (A) \(x^{2}=y^{2}+x y \frac{d y}{d x}\) (B) \(x^{2}=y^{2}+3 x y \frac{d y}{d x}\) (C) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\) (D) \(y^{2}=x^{2}-2 x y \frac{d y}{d x}\)
5 step solution
Problem 109
The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is \(\quad\) (C) \(y=x e^{(x-1)}\) (A) \(y=\ln x+x\) (B) \(y=x \ln x+x^{2}\) (D) \(y=x \ln x+x\)
8 step solution
Problem 110
The differential equation of the family of circles with fixed radius 5 units and centre on the line \(y=2\) is (A) \((x-2) y^{\prime 2}=25-(y-2)^{2}\) (B) \((y-2) y^{\prime 2}=25-(y-2)^{2}\) (C) \((y-2)^{2} y^{\prime 2}=25-(y-2)^{2}\) (D) \((x-2)^{2} y^{\prime 2}=25-(y-2)^{2}\)
6 step solution
Problem 111
The differential equation which represents the family of curves \(y=c_{1} e^{c_{2} x}\), where \(c_{1}\) and \(c_{2}\) are arbitrary constants is (A) \(y^{\prime}=y^{2}\) (B) \(y^{\prime \prime}=y^{\prime} \mathrm{y}\) (C) \(y y^{\prime}=y^{\prime}\) (D) \(y y^{\prime}=\left(y^{\prime}\right)^{2}\)
4 step solution
Problem 113
Let \(l\) be the purchase value of an equipment and \(V(t)\) be the value of equipment after it has been used for \(t\) years. The value \(V(t)\) depreciates at a rate given by the differential equation \(\frac{d V(t)}{d t}=k(T-t)\), where \(k>0\) is a constant and \(T\) is the total life in years of the equipment. Then, the scrap value \(V(T)\) of the equipment is (A) \(l-\frac{k T^{2}}{2}\) (B) \(l-\frac{k(T-t)^{2}}{2}\) (C) \(e^{-k T}\) (D) \(T^{2}-\frac{l}{k}\)
7 step solution
Problem 114
The population \(p(t)\) at time \(t\) of a certain mouse species satisfies the differential equation \(\frac{d p(t)}{d t}=0.5 p(t)\) \(-450\) with initial condition \(p(0)=850\), then the value of \(t\) for which \(p(t)=0\) is (A) \(2 \ln 18\) (B) \(\ln 9\) (C) \(\frac{1}{2} \ln 18\) (D) \(\ln 18\)
6 step solution
Problem 115
At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production \(P\) with respect to additional number of workers \(x\) is given by \(\frac{d P}{d x}=100-12 \sqrt{x}\). If the firm employs 25 more workers, then the new level of production of items is (A) 3000 (B) 3500 (C) 4500 (D) 2500
5 step solution
Problem 116
Let the population of rabbits surviving at a time \(t\) be governed by the differential equation \(\frac{d p(t)}{d t}=\frac{1}{2} p(t)-200\). If initially \(p(0)=100\), then \(p(t)\) equals (A) \(400-300 e^{t / 2}\) (B) \(300-200 e^{-d 2}\) (C) \(600-500 e^{t / 2}\) (D) \(400-300 e^{-t / 2}\)
7 step solution
Problem 117
Let \(y(x)\) be the solution of the differential equation \((x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)\). Then \(y(e)\) is equal to: (A) 0 (B) 2 (C) \(2 e\) (D) \(e\)
7 step solution
Problem 118
If a curve \(y=f(x)\) passes through the point \((1,-1)\) and satisfies the differential equation, \(y(1+x y) d x=x d y\). then \(f\left(-\frac{1}{2}\right)\) is equal to: (A) \(\frac{4}{5}\) (B) \(-\frac{2}{5}\) (C) \(-\frac{4}{5}\) (D) \(\frac{2}{5}\)
5 step solution