12th Standard Mathematics — Ordinary Differential Equations: Online Practice Test

Explanation.

The order is read off directly: the highest derivative present is \( \dfrac{d^2y}{dx^2} \), so the order is \(2\).

For the degree the equation must first be polynomial in its derivatives. The fractional power sits on \( \dfrac{dy}{dx} \), so move that term aside and cube to remove the cube root:

\[ \dfrac{d^2y}{dx^2} + x^{1/4} = -\left(\dfrac{dy}{dx}\right)^{1/3} \;\Longrightarrow\; \left(\dfrac{d^2y}{dx^2} + x^{1/4}\right)^{3} = -\dfrac{dy}{dx}. \]

Now the highest-order derivative appears to the power \(3\), so the degree is \(3\). The fractional power of \(x\) is irrelevant, because degree counts only powers of derivatives. Hence order \(=2\), degree \(=3\).

Explanation.

There are two arbitrary parameters, \(A\) and \(B\), so differentiate twice to eliminate them. From \( y = A\cos(x+B) \):

\[ \dfrac{dy}{dx} = -A\sin(x+B), \qquad \dfrac{d^2y}{dx^2} = -A\cos(x+B). \]

The second derivative equals \(-y\), since \( A\cos(x+B) = y \). Substituting gives \( \dfrac{d^2y}{dx^2} = -y \), that is \( \dfrac{d^2y}{dx^2} + y = 0 \). Both constants drop out automatically, which is why a two-parameter family yields a second-order equation.

Explanation.

Group the \(dx\) and \(dy\) terms so that \( \dfrac{dy}{dx} \) appears:

\[ \left(\sqrt{\sin x} + \sqrt{\cos x}\right)dy = \left(\sqrt{\cos x} - \sqrt{\sin x}\right)dx, \]\[ \dfrac{dy}{dx} = \dfrac{\sqrt{\cos x} - \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}. \]

Only the first derivative appears, so the order is \(1\). The square roots act on \( \sin x \) and \( \cos x \), which are coefficients — not derivatives — so \( \dfrac{dy}{dx} \) occurs to the first power. The degree is therefore \(1\). The trap is to let the square roots suggest degree \(2\); they never touch the derivative.

Explanation.

A general circle is \( (x-h)^2 + (y-k)^2 = a^2 \). Here \(h\), \(k\) and \(a\) are all free to vary, so the family carries three independent arbitrary constants. Each differentiation removes one constant, so three differentiations are needed to eliminate all of them, giving an equation of order \(3\). The rule of thumb is simply: order of the differential equation \(=\) number of independent arbitrary constants. A frequent error is to count only \(h\) and \(k\) and answer \(2\); the radius \(a\) is also free, so the correct order is \(3\).

Explanation.

Two arbitrary constants call for two differentiations. From \( y = Ae^{x} + Be^{-x} \):

\[ \dfrac{dy}{dx} = Ae^{x} - Be^{-x}, \qquad \dfrac{d^2y}{dx^2} = Ae^{x} + Be^{-x}. \]

The second derivative reproduces the original right-hand side, so \( \dfrac{d^2y}{dx^2} = y \), that is \( \dfrac{d^2y}{dx^2} - y = 0 \). (Contrast this with \( y = A\cos x + B\sin x \), which gives \( y'' + y = 0 \); the sign flips because exponentials are their own second derivative.)

Explanation.

This is variables separable. Collect \(y\) on the left and \(x\) on the right:

\[ \dfrac{dy}{y} = \dfrac{dx}{x}. \]

Integrating, \( \ln|y| = \ln|x| + \ln k \), where the single constant is written as \( \ln k \) for convenience. Combining logarithms gives \( \ln|y| = \ln|kx| \), hence \( y = kx \). The solution curves are the straight lines through the origin.

Explanation.

Rearrange and separate the variables:

\[ 2x\,\dfrac{dy}{dx} = y + 3 \;\Longrightarrow\; \dfrac{2\,dy}{y+3} = \dfrac{dx}{x}. \]

Integrating, \( 2\ln|y+3| = \ln|x| + \ln c \), so \( \ln (y+3)^2 = \ln(cx) \) and therefore \( (y+3)^2 = cx \). Writing \( Y = y+3 \), this is \( Y^2 = cx \) — the standard form of a parabola. Hence the solution curves are parabolas.

Explanation.

With a zero right-hand side this is both separable and a homogeneous linear equation. Separating,

\[ \dfrac{dy}{y} = -p(x)\,dx. \]

Integrating gives \( \ln|y| = -\displaystyle\int p(x)\,dx + \ln c \), so \( y = c\,e^{-\int p(x)\,dx} \). Equivalently, the integrating factor is \( e^{\int p\,dx} \) and, since \( Q = 0 \), the linear solution \( y\,e^{\int p\,dx} = c \) gives the same result.

Explanation.

First write the equation in standard linear form \( \dfrac{dy}{dx} + Py = Q \). Splitting the right side,

\[ \dfrac{dy}{dx} + y = \dfrac{1}{x} + \dfrac{y}{x} \;\Longrightarrow\; \dfrac{dy}{dx} + \left(1 - \dfrac{1}{x}\right)y = \dfrac{1}{x}. \]

So \( P = 1 - \dfrac{1}{x} \), and the integrating factor is

\[ \text{I.F.} = e^{\int\left(1 - \frac{1}{x}\right)dx} = e^{\,x - \ln x} = e^{x}\cdot e^{-\ln x} = \dfrac{e^{x}}{x}. \]

The trap is to forget to move the \( \dfrac{y}{x} \) term to the left before reading off \(P\).

Explanation.

By definition the integrating factor of \( \dfrac{dy}{dx} + P(x)\,y = Q(x) \) is \( e^{\int P\,dx} \). We are told this equals \(x\):

\[ e^{\int P\,dx} = x \;\Longrightarrow\; \int P\,dx = \ln x. \]

Differentiating both sides with respect to \(x\) recovers \(P\): \( P = \dfrac{d}{dx}(\ln x) = \dfrac{1}{x} \). So \( P(x) = \dfrac{1}{x} \).

Explanation.

Read the right-hand side as a power series. The pattern \( 1 + t + \dfrac{t^{2}}{2!} + \dfrac{t^{3}}{3!} + \cdots \) is exactly the Maclaurin expansion of \( e^{t} \), with \( t = \dfrac{dy}{dx} \). So the equation is simply

\[ y = e^{\,dy/dx} \;\Longrightarrow\; \dfrac{dy}{dx} = \ln y. \]

This contains only the first derivative, to the first power, so both the order and the degree are \(1\). The infinite series is a disguise; once summed, the equation is first-order and first-degree.

Explanation.

The highest derivative appearing is \( \dfrac{d^{2}y}{dx^{2}} \), so the order is \( p = 2 \). The equation is already polynomial in its derivatives, and that highest derivative occurs to the first power, so the degree is \( q = 1 \). Comparing, \( p = 2 > 1 = q \), i.e. \( p > q \). The product \( y\,\dfrac{dy}{dx} \) and the term \( \cos x \) raise neither the order nor the degree, because both look only at the highest-order derivative.

Explanation.

Isolate the derivative and integrate directly:

\[ \dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1-x^{2}}} \;\Longrightarrow\; dy = -\dfrac{dx}{\sqrt{1-x^{2}}}. \]

Since \( \displaystyle\int \dfrac{dx}{\sqrt{1-x^{2}}} = \sin^{-1}x \), integrating gives \( y = -\sin^{-1}x + c \), that is \( y + \sin^{-1}x = c \).

Explanation.

Separate the variables:

\[ \dfrac{dy}{y} = 2x\,dx. \]

Integrating, \( \ln|y| = x^{2} + c_{1} \). Exponentiating, \( y = e^{x^{2} + c_{1}} = e^{c_{1}}\,e^{x^{2}} \). Renaming the constant \( e^{c_{1}} = c \) gives \( y = c\,e^{x^{2}} \).

Explanation.

Undo the logarithm first: \( \dfrac{dy}{dx} = e^{x+y} = e^{x}\,e^{y} \). Now separate, sending the \(y\)-terms to the left:

\[ e^{-y}\,dy = e^{x}\,dx. \]

Integrating, \( -e^{-y} = e^{x} + c_{1} \). Multiplying by \(-1\) and renaming the constant gives \( e^{x} + e^{-y} = c \).

Explanation.

Write \( 2^{\,y-x} = \dfrac{2^{y}}{2^{x}} \) and separate:

\[ \dfrac{dy}{2^{y}} = \dfrac{dx}{2^{x}} \;\Longrightarrow\; 2^{-y}\,dy = 2^{-x}\,dx. \]

Using \( \displaystyle\int a^{u}\,du = \dfrac{a^{u}}{\ln a} \), integration gives \( \dfrac{2^{-y}}{-\ln 2} = \dfrac{2^{-x}}{-\ln 2} + c_{1} \). Multiplying by \( -\ln 2 \) and tidying the constant, \( 2^{-x} - 2^{-y} = c \), i.e. \( \dfrac{1}{2^{x}} - \dfrac{1}{2^{y}} = c \).

Explanation.

The right-hand side depends only on \( \dfrac{y}{x} \), so the equation is homogeneous. Substitute \( y = vx \), giving \( \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \). The equation becomes

\[ v + x\dfrac{dv}{dx} = v + \dfrac{\varphi(v)}{\varphi'(v)} \;\Longrightarrow\; \dfrac{\varphi'(v)}{\varphi(v)}\,dv = \dfrac{dx}{x}. \]

The left side is \( \dfrac{d}{dv}\ln\varphi(v) \), so integrating gives \( \ln\varphi(v) = \ln x + \ln k \), hence \( \varphi(v) = kx \). Restoring \( v = \dfrac{y}{x} \), the solution is \( \varphi\!\left(\dfrac{y}{x}\right) = kx \).

Explanation.

For \( \dfrac{dy}{dx} + Py = Q \) the integrating factor is \( e^{\int P\,dx} \). Setting this equal to \( \sin x \):

\[ e^{\int P\,dx} = \sin x \;\Longrightarrow\; \int P\,dx = \ln(\sin x). \]

Differentiate to recover \(P\): \( P = \dfrac{d}{dx}\ln(\sin x) = \dfrac{\cos x}{\sin x} = \cot x \).

Explanation.

A key fact: the general solution of a differential equation of order \(m\) contains exactly \(m\) independent arbitrary constants. Applying this, an equation of order \(n\) has \(n\) arbitrary constants and one of order \(n+1\) has \(n+1\). So the answer is \( n,\ n+1 \).

Explanation.

A general solution of a third-order equation carries three arbitrary constants. A particular solution is obtained by fixing those constants with initial or boundary conditions, so none remain free. Hence the number of arbitrary constants in a particular solution is \(0\), whatever the order.

Explanation.

Rewrite the equation in linear form by splitting the right side:

\[ \dfrac{dy}{dx} = \dfrac{(x+1) + y}{x+1} = 1 + \dfrac{y}{x+1} \;\Longrightarrow\; \dfrac{dy}{dx} - \dfrac{1}{x+1}\,y = 1. \]

So \( P = -\dfrac{1}{x+1} \), and the integrating factor is

\[ \text{I.F.} = e^{\int -\frac{1}{x+1}\,dx} = e^{-\ln(x+1)} = \dfrac{1}{x+1}. \]

Explanation.

"Rate of increase proportional to the population" translates to \( \dfrac{dP}{dt} = kP \) with \( k > 0 \). Separating, \( \dfrac{dP}{P} = k\,dt \), and integrating gives \( \ln P = kt + c_{1} \). Exponentiating, \( P = e^{kt + c_{1}} = c\,e^{kt} \). This is the exponential growth law.

Explanation.

Evaporation removes the substance, so the amount decreases at a rate proportional to what remains: \( \dfrac{dP}{dt} = -kP \) with \( k > 0 \). Separating and integrating, \( \ln P = -kt + c_{1} \), hence \( P = c\,e^{-kt} \). The negative exponent marks exponential decay, distinguishing it from growth.

Explanation.

Cross-multiply and integrate:

\[ (2y+f)\,dy = (ax+3)\,dx \;\Longrightarrow\; y^{2} + fy = \dfrac{a}{2}x^{2} + 3x + c. \]

Bringing everything to one side gives \( \dfrac{a}{2}x^{2} - y^{2} + 3x - fy + c = 0 \). A second-degree curve is a circle only when the coefficients of \( x^{2} \) and \( y^{2} \) are equal. Here the coefficient of \( y^{2} \) is \(-1\), so we need \( \dfrac{a}{2} = -1 \), giving \( a = -2 \).

Explanation.

Integrate the slope: \( dy = 3x^{2}\,dx \) gives \( y = x^{3} + c \). Use the point \( (-1, 1) \) to fix \(c\): \( 1 = (-1)^{3} + c = -1 + c \), so \( c = 2 \). The curve is \( y = x^{3} + 2 \).