Differential Equations — JEE MCQs with answers

Answer: C — For $y = \sqrt{1 + x^2}$: L.H.S. $= y' = \frac{x}{\sqrt{1 + x^2}}$ and R.H.S. $= \frac{x\sqrt{1 + x^2}}{1 + x^2} = \frac{x}{\sqrt{1 + x^2}}$. They are equal, so the answer is option C. Option D: for $y = 1 + x^2$, L.H.S. $= 2x$ but R.H.S. $= x$. Option A: for $y = \frac{1}{\sqrt{1 + x^2}}$, L.H.S. $= -\frac{x}{(1 + x^2)^{3/2}}$ but R.H.S. $= +\frac{x}{(1 + x^2)^{3/2}}$. Option B: for $y = x\sqrt{1 + x^2}$, L.H.S. $= \sqrt{1 + x^2} + \frac{x^2}{\sqrt{1 + x^2}}$ but R.H.S. $= \frac{x^2}{\sqrt{1 + x^2}}$. In each of these the two sides differ.

Answer: A — Put $\tan x = t$, so $\sec^{2}x\,dx = dt$ and the integral becomes $\int t\,e^{t}\,dt$. By parts, $\int t\,e^{t}\,dt = t\,e^{t} - \int e^{t}\,dt = e^{t}(t - 1)$. So the integral is $e^{\tan x}(\tan x - 1)$, which leads to the printed answer $y = (\tan x - 1) + Ce^{-\tan x}$ of EXERCISE 9.5, Question 5. The answer is option A. Option D adds $\int e^{t}\,dt$ instead of subtracting it; its derivative is $(t + 2)e^{t}\sec^{2}x$. Option B stops at $t\,e^{t}$ and leaves out the second term; its derivative is $(t + 1)e^{t}\sec^{2}x$. Option C integrates the factor $t$ by itself and keeps $e^{t}$; its derivative is $\left(t + \frac{t^{2}}{2}\right)e^{t}\sec^{2}x$.

Answer: A — The functions $e^{x}$ and $\sin y$ contain no derivative, so this is a polynomial equation in $y''$ and $y'$ and its degree is defined, as for equation (10) of Section 9.2.2. The highest order derivative is $y''$ and it is raised to the power 5, so the degree is 5. The answer is option A. Option B is the order of the equation, not its degree. Option C is the power on $y'$, which is not the highest order derivative. Option D treats $e^{x}$ and $\sin y$ as obstacles, although neither contains a derivative.

Answer: A — The Introduction says: "some methods to solve a first order - first degree differential equation". That is option A. Option B keeps "first order" but drops the restriction on the degree. Option C keeps "first degree" but drops the restriction on the order. Option D changes the order named in the Introduction from first to second.

Answer: B — For $y = e^{kx}$: $\frac{dy}{dx} = ke^{kx}$ and $\frac{d^2y}{dx^2} = k^2e^{kx}$. Substituting, L.H.S. $= (k^2 - 5k + 6)e^{kx} = (k - 2)(k - 3)e^{kx}$. As $e^{kx} \neq 0$, the L.H.S. equals the R.H.S. $0$ only for $k = 2$ or $k = 3$. The answer is option B. Option A has both signs wrong: $k = -2$ gives $(4 + 10 + 6)e^{-2x} = 20e^{-2x}$ and $k = -3$ gives $30e^{-3x}$. Option C uses factors of 6 that do not add up to 5: $k = 1$ gives $2e^{x}$ and $k = 6$ gives $12e^{6x}$. Option D factorises $k^2 - 5k - 6$ instead: $k = -1$ gives $12e^{-x}$ and $k = 6$ gives $12e^{6x}$.

Answer: B — Write $\frac{2x^{2} + x}{(x + 1)(x^{2} + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^{2} + 1}$. Then $2x^{2} + x = A(x^{2} + 1) + (Bx + C)(x + 1)$. Putting $x = -1$ gives $1 = 2A$, so $A = \frac{1}{2}$; comparing $x^{2}$ terms, $A + B = 2$, so $B = \frac{3}{2}$; comparing constants, $A + C = 0$, so $C = -\frac{1}{2}$. Integrating this form gives $\frac{1}{2}\log(x + 1) + \frac{3}{4}\log(x^{2} + 1) - \frac{1}{2}\tan^{-1}x$, which is the printed answer's $\frac{1}{4}\log\left[(x + 1)^{2}(x^{2} + 1)^{3}\right] - \frac{1}{2}\tan^{-1}x$. The answer is option B. Option A has the wrong sign on the constant in the second numerator; its numerator adds up to $2x^{2} + 2x + 1$. Option C adds up to $\frac{2x^{2}}{(x + 1)(x^{2} + 1)}$, so the term $x$ is lost. Option D interchanges the coefficients $\frac{1}{2}$ and $\frac{3}{2}$; its numerator adds up to $2x^{2} + 1$.

Answer: B — The derivatives involved are $y'''$ and $y'$, and the highest order derivative is $y''' = \dfrac{d^{3}y}{dx^{3}}$, so the order is 3. That $y'''$ sits in the exponent of $e$ makes no difference to the order: the definition of order only asks for the highest order derivative involved. The answer is option B. Option A: this is the order of $y'$, which is not the highest order derivative involved. Option C: this is the power carried by $y'$. Option D: the order is defined for every differential equation; it is the degree that needs the equation to be a polynomial equation in its derivatives.

Answer: B — The Historical Note says the identity $\int y\,dy = \frac{1}{2}y^{2}$ was written down "thereby introducing both the symbols $\int$ and $dy$". That is option B. Option A replaces $dy$ by $dx$, which does not occur in the identity. Option C leaves out the integral sign. Option D names the dash notation for derivatives, which the Historical Note does not mention.

Answer: D — Divide by $1 - y^{2}$: $\dfrac{dx}{dy} + \dfrac{y}{1 - y^{2}}x = \dfrac{ay}{1 - y^{2}}$, so $P_1 = \dfrac{y}{1 - y^{2}}$. With $t = 1 - y^{2}$, $dt = -2y\,dy$, so $\int P_1\,dy = -\dfrac{1}{2}\log(1 - y^{2})$ (and $1 - y^{2} > 0$ on $-1 < y < 1$). Hence $e^{\int P_1\,dy} = (1 - y^{2})^{-1/2} = \dfrac{1}{\sqrt{1 - y^{2}}}$, option (D), the printed answer. Option (C) drops the factor $\dfrac{1}{2}$; option (A) is its negative; option (B) is not even real on $-1 < y < 1$, where $y^{2} - 1 < 0$.

Answer: B — According to the Summary, the general solution contains as many arbitrary constants as the order of the differential equation. The highest order derivative in the equation is $\frac{d^2y}{dx^2}$, so the order is $2$. Hence the general solution contains $2$ arbitrary constants. The answer is option B. Option C takes the power $3$ on $\frac{d^2y}{dx^2}$, which is the degree, not the order. Option D adds the order $2$ and the degree $3$. Option A takes the order of the derivative $\frac{dy}{dx}$, which is not the highest order derivative.