Chapter 17

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

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

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) \(\frac{x}{y}=\ln \left(e^{x y}+c\right)\)

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) \(y=\frac{c}{x} e^{\frac{1}{x^{\prime}}}\)

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

The equation of the curve satisfying the differential equation \(\sqrt{x-y \frac{d y}{d x}}=\left|x^{2}-y^{2}\right|\) and passing through the point \((1,0)\) is (A) \((2 x-1)+\frac{1}{x^{2}-y^{2}}=0\) (B) \((3 x-2)+\frac{1}{x^{2}-y^{2}}=0\) (C) \((2 x-3)+\frac{1}{x^{2}-y^{2}}=0\) (D) None of these

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) \(y^{2} \sin x=4 \cos ^{2} x\)

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

The solution of the equation \(y e^{-x / y} d x-\left(x e^{-x / y}+y^{3}\right)\) \(d y=0\) is (A) \(3 e^{-x / y}+y^{2}=c\) (B) \(2 e^{-x / y}+y^{2}=c\) (C) \(2 e^{-x / y}+y^{2}=c\) (D) None of these

The differential equation corresponding to \(y=\sum_{i=1}^{3} c_{i} e^{m, x}\), where \(c_{i}\) 's are arbitrary constants and \(m_{1}, m_{2}, m_{3}\) are roots of the equation \(m^{3}-7 m+6=0\), is (A) \(y_{3}-7 y_{1}+6 y=0\) (B) \(y_{3}+7 y_{1}+6 y=0\) (C) \(y_{3}-7 y_{1}-6 y=0\) (D) None of these

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) \(-\frac{1}{2}\)

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

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

A differential equation is said to be exact if it can be derived from its primitive by direct differentiation without any further transformation such as elimination, etc. The differential equation $$ \left(x^{2}-a y\right) d x+\left(y^{2}-a x\right) d y=0 $$ is exact in as much as it can be derived from its primitive $$ x^{3}-3 a x y+y^{3}=c $$ by direct differentiation. The necessary and sufficient condition for the differential equation \(M d x+N d y=0\) to be exact is that $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$ where \(\frac{\partial M}{\partial y}\) is the derivative of \(M\) with respect to \(y\) keeping \(x\) as constant and \(\frac{\partial N}{\partial x}\) is the derivative of \(N\) with respect to \(x\) keeping \(y\) as constant. If the equation \(M d x+N d y=0\) is exact, then it can be integrated as follows: Firstly, integrate \(M\) with respect to \(x\) regarding \(y\) as constant. Then, integrate with respect to \(y\) those of the terms in \(N\) which do not involve \(x\). The sum of the two expressions thus obtained equated to a constant is the required solution. The solution of the equation \(x d x+y d y=\frac{a^{2}(x d y-y d x)}{x^{2}-y^{2}}\) is (A) \(x^{2}+y^{2}+2 a^{2} \tan ^{-1} \frac{x}{y}=A\) (B) \(x^{2}+y^{2}-2 a^{2} \tan ^{-1} \frac{x}{y}=A\) (C) \(x^{2}-y^{2}+2 a^{2} \tan ^{-1} \frac{x}{y}=A\) (D) None of these

A differential equation is said to be exact if it can be derived from its primitive by direct differentiation without any further transformation such as elimination, etc. The differential equation $$ \left(x^{2}-a y\right) d x+\left(y^{2}-a x\right) d y=0 $$ is exact in as much as it can be derived from its primitive $$ x^{3}-3 a x y+y^{3}=c $$ by direct differentiation. The necessary and sufficient condition for the differential equation \(M d x+N d y=0\) to be exact is that $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$ where \(\frac{\partial M}{\partial y}\) is the derivative of \(M\) with respect to \(y\) keeping \(x\) as constant and \(\frac{\partial N}{\partial x}\) is the derivative of \(N\) with respect to \(x\) keeping \(y\) as constant. If the equation \(M d x+N d y=0\) is exact, then it can be integrated as follows: Firstly, integrate \(M\) with respect to \(x\) regarding \(y\) as constant. Then, integrate with respect to \(y\) those of the terms in \(N\) which do not involve \(x\). The sum of the two expressions thus obtained equated to a constant is the required solution. The solution of the equation \(\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x+[x+\log x-x \sin y] d y=0 i s\) (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

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) 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

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) 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

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) The integrating factor to make the differential equation \(\left(x y^{2}-e^{\frac{1}{x^{x}}}\right) d x-x^{2} y d y=0\) exact is (A) \(\frac{1}{x}\) (B) \(\frac{1}{x^{2}}\) (C) \(\frac{1}{x^{3}}\) (D) \(\frac{1}{x^{4}}\)

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) 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) \(\frac{1}{y^{4}}\)

A differential equation of the form \(y=p x+f(p)\), where \(p=\frac{d y}{d x}\) is known as Clairaut's equation. We now find the solution of the above equation. The given equation is $$ y=p x+f(p) $$ It is solvable for \(y\). Differentiating with respect to \(x\), we get $$ \frac{d y}{d x}=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow\) $$ p=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow \quad x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}=0\) Factorizing \(\frac{d p}{d x}+\left[x+f^{\prime}(p)\right]=0\) Cancelling the factor \(x+f^{\prime}(p)\) which does not involve \(\frac{d p}{d x}\), we have \(\frac{d p}{d x}=0\) Integrating, $$ p=c $$ Eliminating \(p\) between (1) and (2), the required solution of \((1)\) is $$ y=c x+f(c) $$ The solution of the equation \((y-p x)(p-1)=p\) is (A) \(y=c x+\frac{c}{c-1}\) (B) \(y=c x-\frac{c}{c-1}\) (C) \(x=c y+\frac{c}{c-1}\) (D) None of these

A differential equation of the form \(y=p x+f(p)\), where \(p=\frac{d y}{d x}\) is known as Clairaut's equation. We now find the solution of the above equation. The given equation is $$ y=p x+f(p) $$ It is solvable for \(y\). Differentiating with respect to \(x\), we get $$ \frac{d y}{d x}=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow\) $$ p=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow \quad x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}=0\) Factorizing \(\frac{d p}{d x}+\left[x+f^{\prime}(p)\right]=0\) Cancelling the factor \(x+f^{\prime}(p)\) which does not involve \(\frac{d p}{d x}\), we have \(\frac{d p}{d x}=0\) Integrating, $$ p=c $$ Eliminating \(p\) between (1) and (2), the required solution of \((1)\) is $$ y=c x+f(c) $$ The solution of the equation \(\left(\frac{d y}{d x}\right)^{2}\left(x^{2}-a^{2}\right)-2\left(\frac{d y}{d x}\right) x y+y^{2}-b^{2}=0\) is (A) \(y=c x+\sqrt{a^{2} c^{2}+b^{2}}\) (B) \(y=c x+\sqrt{b^{2} c^{2}+a^{2}}\) (C) \(y=c x-\sqrt{a^{2} c^{2}+b^{2}}\) (D) \(y=c x-\sqrt{b^{2} c^{2}+a^{2}}\)

A differential equation of the form \(y=p x+f(p)\), where \(p=\frac{d y}{d x}\) is known as Clairaut's equation. We now find the solution of the above equation. The given equation is $$ y=p x+f(p) $$ It is solvable for \(y\). Differentiating with respect to \(x\), we get $$ \frac{d y}{d x}=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow\) $$ p=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow \quad x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}=0\) Factorizing \(\frac{d p}{d x}+\left[x+f^{\prime}(p)\right]=0\) Cancelling the factor \(x+f^{\prime}(p)\) which does not involve \(\frac{d p}{d x}\), we have \(\frac{d p}{d x}=0\) Integrating, $$ p=c $$ Eliminating \(p\) between (1) and (2), the required solution of \((1)\) is $$ y=c x+f(c) $$ 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) \((y-c x+2 c)(y-c x-1)=0\)

A curve which cuts every member of a given family of curves according to a given law is called a trajectory of the family. We shall consider only the case when each trajectory cuts every member of a given family at a constant angle. The trajectory will be called Orthogonal if the constant angle is a right angle. For example, every line through the origin of coordinates is an orthogonal trajectory of the family of concentric circles with centre at the origin. Let the equation of the given family of curves be $$ f(x, y, c)=0 $$ Differentiate (1) and eliminate the arbitaray constant \(c\) between (1) and the resulting equation. That gives the differential equation of the family (1). Let it be \(F\left(x, y, \frac{d y}{d x}\right)=0\) Replace \(\frac{d y}{d x}\) by \(-\frac{d x}{d y}\) The differential equation of the orthogonal trajectory is $$ F\left(x, y,-\frac{d x}{d y}\right)=0 $$ Integrate (3) to get the equation of the required orthogonal trajectory. Orthogonal trajectory of the family of hyperbolas \(x y=k^{2}\) is (A) \(x^{2}+y^{2}=c\) (B) \(x^{2}-y^{2}=c\) (C) \(2 x^{2}-2 y^{2}=c\) (D) None of these

The differential equation of family of Column-I I. Circles passing through the origin and having their centres on the \(x\)-axis II. Parabolas with foci at the origin and axis along the \(x\)-axis III. Parabolas each of which has a latus rectum \(4 a\) and whose axes are parallel to \(x\)-axis IV. Ellipsescentred at the origin Column-II (A) \(x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\) (B) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\) (C) \(y\left(\frac{d y}{d x}\right)^{2}+2 x\left(\frac{d y}{d x}\right)-y=0\) (D) \(2 a \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}=0\)

Column-I I. \(y d x-x d y+(\log x) d x=0\) II. \(\left(x^{2} \sin ^{3} y-y^{2} \cos x\right) d x+\left(x^{3} \cos \right.\) \(\left.y \sin ^{2} y-2 y \sin x\right) d y=0\) III. \(\frac{d y}{d x}=\frac{y(x \log y-y)}{x(y \log x-x)}\) IV. \(\frac{x d y}{x^{2}+y^{2}}=\left(\frac{y}{x^{2}+y^{2}}-1\right) d x\) Column-II (A) \(\tan ^{-1} \frac{y}{x}+x=c\) (B) \(y=1+\log x+c\) (C) \(x^{y}=c y^{x}\) (D) \(\frac{x^{3} \sin ^{3} y}{3}\) \(=y^{2} \sin x+c\)

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is acorrect explanation forAssertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True 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 \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0\)

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is acorrect explanation forAssertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: The solution of the equation \(x \sin \theta d \theta+\) \(\left(x^{3}-2 x^{2} \cos \theta+\cos \theta\right) d x=0\) is \(2 \cos \theta=x+c x e^{-x^{2}}\) Reason: Integrating factor \(=\frac{e^{x^{2}}}{x}\)

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is acorrect explanation forAssertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True 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 \(x \cos \alpha+y \sin \alpha=p .\)

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 \(\quad[\mathbf{2 0 0 2}]\) (A) \(\left(1, \frac{2}{3}\right)\) (B) \((3,1)\) (C) \((3,3)\) (D) \((1,2)\)

The solution of the equation \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) is [2002] (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) \(\frac{1}{4} e^{-2 x}+c+d\)

The differential equation of all non-vertical lines in a plane is [2002] (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) \(\frac{d x}{d y}=0\)

The degree and order of the differential equation of the family of all parabolas whose axis is \(x\)-axis, are respectively (A) 2,1 (B) 1,2 (C) 3,2 (D) 2,3

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 \(\quad[\mathbf{2 0 0 3}]\) (A) \((x-2)=k e^{-\tan ^{-1} y}\) (B) \(2 x e^{\tan ^{-1}} y=e^{2 \tan ^{-1}}+k\) (C) \(x e^{\tan ^{-1} y}=\tan ^{-1} y+k\) (D) \(x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+k\)

The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant is [2004] (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) \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)

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) \(\log y=C x\)

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: \([2005]\) (A) order 1, degree 2 (B) order 1, degree 1 (C) order 1, degree 3 (D) order 2, degree 2

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) \(\log \left(\frac{x}{y}\right)=c y\)

The differential equation whose solution is \(A x^{2}+B y^{2}\) \(=1\), where \(A\) and \(B\) are arbitrary constants is of [2006] (A) second order and second degree (B) first order and second degree (C) first order and first degree (D) second order and first degree

The differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is (A) \(x^{2}=y^{2}+x y \frac{d y}{d x}\) [2007] (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) \(y^{2}=x^{2}-2 x y \frac{d y}{d x}\)

The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is \([2008]\) (A) \(y=\ln x+x\) (B) \(y=x \ln x+x^{2}\) (C) \(y=x e^{(x-1)}\) (D) \(y=x \ln x+x\)

The differential equation of the family of circles with fixed radius 5 units and centre on the line \(y=2\) is [2008] (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) \((x-2)^{2} y^{\prime 2}=25-(y-2)^{2}\)

The differential equation which represents the family of curves \(y=c_{1} e^{c_{x} x}\), where \(c_{1}\) and \(c_{2}\) are arbitrary constants is (A) \(y^{\prime}=y^{2}\) (B) \(y^{\prime \prime}=y^{\prime} y\) (C) \(y y^{\prime}=y^{\prime}\) (D) \(y y^{\prime}=\left(y^{\prime}\right)^{2}\)

Solution to the differential equation \(\cos x d y=y(\sin\) \(x-y) d x, 0

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 [2011] (A) \(l-\frac{k T^{2}}{2}\) (B) \(l-\frac{k(T-t)^{2}}{2}\) (C) \(e^{-k T}\) (D) \(T^{2}-\frac{l}{k}\)

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 \([2012]\) (A) \(2 \ln 18\) (B) \(\ln 9\) (C) \(\frac{1}{2} \ln 18\) (D) \(\ln 18\)

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 [2013] (A) 3000 (B) 3500 (C) 4500 (D) 2500

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 [2014] (A) \(400-300 e^{[/ 2}\) (B) \(300-200 e^{-t / 2}\) (C) \(600-500 e^{t / 2}\) (D) \(400-300 e^{-t / 2}\)

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: [2015] (A) 0 (B) 2 (C) \(2 e\) (D) \(e\)

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) \(\frac{2}{5}\)