Differential Equations — NCERT Solutions

Given: $\dfrac{d^4y}{dx^4} + \sin(y''') = 0$

Order: The highest order derivative present is $\dfrac{d^4y}{dx^4}$ (i.e., the 4th derivative), so the order is 4.

Degree: The equation contains $\sin(y''')$, which is a transcendental (non-polynomial) function of the derivative $y'''$. Therefore, the equation is not a polynomial in its derivatives.

Degree is not defined.

Given: $y' + 5y = 0$

Order: The highest order derivative present is $y' = \dfrac{dy}{dx}$, so the order is 1.

Degree: The equation is a polynomial in $y'$ and the highest power of $y'$ is 1.

Degree is 1.

Given: $\left(\dfrac{ds}{dt}\right)^4 + 3s\dfrac{d^2s}{dt^2} = 0$

Order: The highest order derivative present is $\dfrac{d^2s}{dt^2}$, so the order is 2.

Degree: The equation is a polynomial in $\dfrac{d^2s}{dt^2}$ and $\dfrac{ds}{dt}$. The highest power of the highest order derivative $\dfrac{d^2s}{dt^2}$ is 1.

Degree is 1.

Given: $\left(\dfrac{d^2y}{dx^2}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) = 0$

Order: The highest order derivative present is $\dfrac{d^2y}{dx^2}$, so the order is 2.

Degree: The equation contains $\cos\left(\dfrac{dy}{dx}\right)$, which is a transcendental function of the derivative $\dfrac{dy}{dx}$. Therefore, the equation is not a polynomial in its derivatives.

Degree is not defined.

Given: $\dfrac{d^2y}{dx^2} = \cos 3x + \sin 3x$

Order: The highest order derivative present is $\dfrac{d^2y}{dx^2}$, so the order is 2.

Degree: The equation is a polynomial in $\dfrac{d^2y}{dx^2}$ and the highest power of $\dfrac{d^2y}{dx^2}$ is 1.

Degree is 1.

Given: $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$

Order: The highest order derivative present is $y'''$ (3rd derivative), so the order is 3.

Degree: The equation is a polynomial in $y'''$, $y''$, $y'$. The highest power of the highest order derivative $y'''$ is 2.

Degree is 2.

Given: $y''' + 2y'' + y' = 0$

Order: The highest order derivative present is $y'''$ (3rd derivative), so the order is 3.

Degree: The equation is a polynomial in $y'''$, $y''$, $y'$ and the highest power of $y'''$ is 1.

Degree is 1.

Given: $y' + y = e^x$

Order: The highest order derivative present is $y' = \dfrac{dy}{dx}$, so the order is 1.

Degree: The equation is a polynomial in $y'$ and the highest power of $y'$ is 1.

Degree is 1.

Given: $y'' + (y')^2 + 2y = 0$

Order: The highest order derivative present is $y''$ (2nd derivative), so the order is 2.

Degree: The equation is a polynomial in $y''$ and $y'$. The highest power of the highest order derivative $y''$ is 1.

Degree is 1.

Given: $y'' + 2y' + \sin y = 0$

Order: The highest order derivative present is $y''$ (2nd derivative), so the order is 2.

Degree: The equation is a polynomial in $y''$ and $y'$. The term $\sin y$ involves $y$ (the dependent variable), not a derivative, so it does not affect the polynomial nature in derivatives. The highest power of $y''$ is 1.

Degree is 1.

Correct Option: (D) not defined

Justification: The equation contains the term $\sin\left(\dfrac{dy}{dx}\right)$, which is a transcendental (non-polynomial) function of the derivative $\dfrac{dy}{dx}$. Since the equation is not a polynomial in its derivatives, the degree is not defined.

Correct Option: (A) 2

Justification: The highest order derivative present in the equation is $\dfrac{d^2y}{dx^2}$ (2nd order derivative). Therefore, the order of the differential equation is 2.

Given: $y = e^x + 1$

Differentiating once: $y' = e^x$

Differentiating again: $y'' = e^x$

Substituting in the differential equation:
$$y'' - y' = e^x - e^x = 0 = \text{R.H.S.}$$

Hence, $y = e^x + 1$ is a solution of $y'' - y' = 0$. ✓

Given: $y = x^2 + 2x + C$

Differentiating: $y' = 2x + 2$

Substituting in the differential equation:
$$y' - 2x - 2 = (2x + 2) - 2x - 2 = 0 = \text{R.H.S.}$$

Hence, $y = x^2 + 2x + C$ is a solution of $y' - 2x - 2 = 0$. ✓

Given: $y = \cos x + C$

Differentiating: $y' = -\sin x$

Substituting in the differential equation:
$$y' + \sin x = -\sin x + \sin x = 0 = \text{R.H.S.}$$

Hence, $y = \cos x + C$ is a solution of $y' + \sin x = 0$. ✓

Given: $y = \sqrt{1+x^2} = (1+x^2)^{1/2}$

Differentiating:
$$y' = \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$$

R.H.S.:
$$\frac{xy}{1+x^2} = \frac{x \cdot \sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$$

Since L.H.S. $=$ R.H.S., $y = \sqrt{1+x^2}$ is a solution. ✓

Given: $y = Ax$

Differentiating: $y' = A$

Substituting in the differential equation:
$$xy' = x \cdot A = Ax = y = \text{R.H.S.}$$

Hence, $y = Ax$ is a solution of $xy' = y$. ✓

Given: $y = x\sin x$

Differentiating: $y' = \sin x + x\cos x$

L.H.S.: $xy' = x(\sin x + x\cos x) = x\sin x + x^2\cos x$

R.H.S.: $y + x\sqrt{x^2 - y^2}$

Now, $y^2 = x^2\sin^2 x$, so:
$$x^2 - y^2 = x^2 - x^2\sin^2 x = x^2(1 - \sin^2 x) = x^2\cos^2 x$$
$$\sqrt{x^2 - y^2} = |x\cos x| = x\cos x \quad (\text{assuming } x\cos x > 0)$$

Therefore:
$$y + x\sqrt{x^2 - y^2} = x\sin x + x \cdot x\cos x = x\sin x + x^2\cos x$$

L.H.S. $=$ R.H.S. ✓

Given: $xy = \log y + C$

Differentiating both sides with respect to $x$:
$$y + x\frac{dy}{dx} = \frac{1}{y}\frac{dy}{dx}$$

$$y = \frac{dy}{dx}\left(\frac{1}{y} - x\right) = \frac{dy}{dx}\cdot\frac{1-xy}{y}$$

$$\frac{dy}{dx} = \frac{y^2}{1-xy}$$

This is exactly the given differential equation. Hence verified. ✓

Given: $y - \cos y = x$

Differentiating both sides with respect to $x$:
$$\frac{dy}{dx} + \sin y \cdot \frac{dy}{dx} = 1$$
$$\frac{dy}{dx}(1 + \sin y) = 1$$
$$\frac{dy}{dx} = \frac{1}{1 + \sin y}$$

Now check L.H.S. of the differential equation:
$$(y\sin y + \cos y + x)y'$$

Since $x = y - \cos y$:
$$y\sin y + \cos y + x = y\sin y + \cos y + y - \cos y = y\sin y + y = y(1 + \sin y)$$

Therefore:
$$(y\sin y + \cos y + x)y' = y(1+\sin y) \cdot \frac{1}{1+\sin y} = y = \text{R.H.S.}$$

Hence verified. ✓

Given: $x + y = \tan^{-1}y$

Differentiating both sides with respect to $x$:
$$1 + \frac{dy}{dx} = \frac{1}{1+y^2}\cdot\frac{dy}{dx}$$

$$1 = \frac{dy}{dx}\left(\frac{1}{1+y^2} - 1\right) = \frac{dy}{dx}\cdot\frac{1 - (1+y^2)}{1+y^2} = \frac{dy}{dx}\cdot\frac{-y^2}{1+y^2}$$

$$\frac{dy}{dx} = \frac{-(1+y^2)}{y^2}$$

Substituting in the differential equation $y^2y' + y^2 + 1 = 0$:
$$y^2 \cdot \frac{-(1+y^2)}{y^2} + y^2 + 1 = -(1+y^2) + y^2 + 1 = -1 - y^2 + y^2 + 1 = 0 = \text{R.H.S.}$$

Hence verified. ✓

Given: $y = \sqrt{a^2 - x^2}$

Differentiating:
$$\frac{dy}{dx} = \frac{-2x}{2\sqrt{a^2-x^2}} = \frac{-x}{\sqrt{a^2-x^2}} = \frac{-x}{y}$$

Substituting in the differential equation:
$$x + y\frac{dy}{dx} = x + y \cdot \frac{-x}{y} = x - x = 0 = \text{R.H.S.}$$

Hence verified. ✓

Correct Option: (D) 4

Justification: The general solution of a differential equation of order $n$ contains exactly $n$ arbitrary constants. For a fourth order differential equation, the general solution contains 4 arbitrary constants.

Correct Option: (D) 0

Justification: A particular solution is obtained from the general solution by assigning specific values to all arbitrary constants. Therefore, a particular solution contains no arbitrary constants, regardless of the order of the differential equation.

Given: $\dfrac{dy}{dx} = \dfrac{1-\cos x}{1+\cos x}$

Using identities: $1 - \cos x = 2\sin^2\dfrac{x}{2}$ and $1 + \cos x = 2\cos^2\dfrac{x}{2}$

$$\frac{dy}{dx} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2} = \sec^2\frac{x}{2} - 1$$

Separating variables and integrating:
$$dy = \left(\sec^2\frac{x}{2} - 1\right)dx$$

$$\int dy = \int\left(\sec^2\frac{x}{2} - 1\right)dx$$

$$y = \frac{\tan\frac{x}{2}}{\frac{1}{2}} - x + C$$

$$\boxed{y = 2\tan\frac{x}{2} - x + C}$$

Given: $\dfrac{dy}{dx} = \sqrt{4-y^2}$

Separating variables:
$$\frac{dy}{\sqrt{4-y^2}} = dx$$

Integrating both sides:
$$\int\frac{dy}{\sqrt{4-y^2}} = \int dx$$

$$\sin^{-1}\frac{y}{2} = x + C$$

$$\boxed{\sin^{-1}\frac{y}{2} = x + C} \quad \text{or} \quad y = 2\sin(x+C)$$

Given: $\dfrac{dy}{dx} + y = 1$, i.e., $\dfrac{dy}{dx} = 1 - y$

Separating variables:
$$\frac{dy}{1-y} = dx$$

Integrating both sides:
$$\int\frac{dy}{1-y} = \int dx$$

$$-\log|1-y| = x + C_1$$

$$\log|1-y| = -x - C_1$$

$$|1-y| = e^{-x-C_1} = e^{-C_1}\cdot e^{-x}$$

$$1 - y = \pm e^{-C_1}\cdot e^{-x} = Ae^{-x} \quad (\text{where } A = \pm e^{-C_1})$$

$$\boxed{y = 1 - Ae^{-x}}$$

or equivalently $y = 1 + Ce^{-x}$ where $C$ is an arbitrary constant.

Given: $\sec^2 x \tan y\, dx + \sec^2 y \tan x\, dy = 0$

Separating variables (dividing by $\tan x \tan y$):
$$\frac{\sec^2 x}{\tan x}dx + \frac{\sec^2 y}{\tan y}dy = 0$$

Integrating both sides:
$$\int\frac{\sec^2 x}{\tan x}dx + \int\frac{\sec^2 y}{\tan y}dy = 0$$

Let $\tan x = u \Rightarrow \sec^2 x\, dx = du$ for the first integral, and similarly for the second:
$$\log|\tan x| + \log|\tan y| = \log C$$

$$\log|\tan x \cdot \tan y| = \log C$$

$$\boxed{\tan x \cdot \tan y = C}$$

Given: $(e^x + e^{-x})dy = (e^x - e^{-x})dx$

Separating variables:
$$dy = \frac{e^x - e^{-x}}{e^x + e^{-x}}dx$$

Integrating both sides:
$$\int dy = \int\frac{e^x - e^{-x}}{e^x + e^{-x}}dx$$

Note that the numerator is the derivative of the denominator:
$$y = \log|e^x + e^{-x}| + C$$

$$\boxed{y = \log(e^x + e^{-x}) + C}$$

(Since $e^x + e^{-x} > 0$, absolute value is not needed.)

Given: $\dfrac{dy}{dx} = (1+x^2)(1+y^2)$

Separating variables:
$$\frac{dy}{1+y^2} = (1+x^2)dx$$

Integrating both sides:
$$\int\frac{dy}{1+y^2} = \int(1+x^2)dx$$

$$\tan^{-1}y = x + \frac{x^3}{3} + C$$

$$\boxed{\tan^{-1}y = x + \frac{x^3}{3} + C}$$

Given: $y\log y\, dx = x\, dy$

Separating variables:
$$\frac{dx}{x} = \frac{dy}{y\log y}$$

Integrating both sides:
$$\int\frac{dx}{x} = \int\frac{dy}{y\log y}$$

For the right side, let $\log y = t \Rightarrow \dfrac{1}{y}dy = dt$:
$$\log|x| = \int\frac{dt}{t} = \log|t| = \log|\log y|$$

$$\log|x| = \log|\log y| + \log C$$

$$\log|x| = \log|C\log y|$$

$$\boxed{x = C\log y}$$

Given: $x^5\dfrac{dy}{dx} = -y^5$

Separating variables:
$$\frac{dy}{y^5} = -\frac{dx}{x^5}$$

$$y^{-5}dy = -x^{-5}dx$$

Integrating both sides:
$$\int y^{-5}dy = -\int x^{-5}dx$$

$$\frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C_1$$

$$-\frac{1}{4y^4} = \frac{1}{4x^4} + C_1$$

Multiplying by $-4$:
$$\frac{1}{y^4} = -\frac{1}{x^4} + C \quad (C = -4C_1)$$

$$\boxed{x^{-4} + y^{-4} = C}$$

Given: $\dfrac{dy}{dx} = \sin^{-1}x$

Separating variables:
$$dy = \sin^{-1}x\, dx$$

Integrating both sides:
$$y = \int \sin^{-1}x\, dx$$

Using integration by parts: $u = \sin^{-1}x$, $dv = dx$
$$\frac{du}{dx} = \frac{1}{\sqrt{1-x^2}}, \quad v = x$$

$$y = x\sin^{-1}x - \int\frac{x}{\sqrt{1-x^2}}dx$$

For $\displaystyle\int\frac{x}{\sqrt{1-x^2}}dx$: let $1-x^2 = t \Rightarrow -2x\,dx = dt$
$$\int\frac{x}{\sqrt{1-x^2}}dx = -\frac{1}{2}\int t^{-1/2}dt = -\sqrt{t} = -\sqrt{1-x^2}$$

Therefore:
$$y = x\sin^{-1}x - (-\sqrt{1-x^2}) + C$$

$$\boxed{y = x\sin^{-1}x + \sqrt{1-x^2} + C}$$

Given: $e^x\tan y\, dx + (1-e^x)\sec^2 y\, dy = 0$

Separating variables (dividing by $\tan y(1-e^x)$):
$$\frac{e^x}{1-e^x}dx + \frac{\sec^2 y}{\tan y}dy = 0$$

Integrating both sides:
$$\int\frac{e^x}{1-e^x}dx + \int\frac{\sec^2 y}{\tan y}dy = 0$$

For the first integral, let $1-e^x = t \Rightarrow -e^x dx = dt$:
$$\int\frac{e^x}{1-e^x}dx = -\int\frac{dt}{t} = -\log|t| = -\log|1-e^x|$$

For the second integral, let $\tan y = u \Rightarrow \sec^2 y\, dy = du$:
$$\int\frac{\sec^2 y}{\tan y}dy = \log|\tan y|$$

Therefore:
$$-\log|1-e^x| + \log|\tan y| = \log C$$

$$\log\left|\frac{\tan y}{1-e^x}\right| = \log C$$

$$\boxed{\tan y = C(1-e^x)}$$

Given: $(x^3+x^2+x+1)\dfrac{dy}{dx} = 2x^2+x$

Separating variables:
$$dy = \frac{2x^2+x}{x^3+x^2+x+1}dx$$

Factoring the denominator:
$$x^3+x^2+x+1 = x^2(x+1)+(x+1) = (x^2+1)(x+1)$$

Partial fractions:
$$\frac{2x^2+x}{(x^2+1)(x+1)} = \frac{A}{x+1} + \frac{Bx+D}{x^2+1}$$

$$2x^2+x = A(x^2+1) + (Bx+D)(x+1)$$

Setting $x = -1$: $2-1 = A(2) \Rightarrow A = \dfrac{1}{2}$

Expanding: $2x^2+x = Ax^2+A+Bx^2+Bx+Dx+D$

Comparing $x^2$: $2 = A+B \Rightarrow B = 2 - \dfrac{1}{2} = \dfrac{3}{2}$

Comparing constant: $0 = A+D \Rightarrow D = -\dfrac{1}{2}$

Comparing $x$: $1 = B+D = \dfrac{3}{2} - \dfrac{1}{2} = 1$ ✓

Integrating:
$$y = \int\left[\frac{1/2}{x+1} + \frac{\frac{3}{2}x - \frac{1}{2}}{x^2+1}\right]dx$$

$$y = \frac{1}{2}\log|x+1| + \frac{3}{4}\log(x^2+1) - \frac{1}{2}\tan^{-1}x + C$$

Applying condition $y=1$ when $x=0$:
$$1 = \frac{1}{2}\log 1 + \frac{3}{4}\log 1 - \frac{1}{2}\cdot 0 + C \Rightarrow C = 1$$

Particular solution:
$$\boxed{y = \frac{1}{2}\log|x+1| + \frac{3}{4}\log(x^2+1) - \frac{1}{2}\tan^{-1}x + 1}$$

Given: $x(x^2-1)\dfrac{dy}{dx} = 1$

Separating variables:
$$dy = \frac{dx}{x(x^2-1)} = \frac{dx}{x(x-1)(x+1)}$$

Partial fractions:
$$\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$$

$x=0$: $1 = A(-1)(1) \Rightarrow A = -1$

$x=1$: $1 = B(1)(2) \Rightarrow B = \dfrac{1}{2}$

$x=-1$: $1 = C(-1)(-2) \Rightarrow C = \dfrac{1}{2}$

Integrating:
$$y = -\log|x| + \frac{1}{2}\log|x-1| + \frac{1}{2}\log|x+1| + K$$

$$y = \frac{1}{2}\log|(x-1)(x+1)| - \log|x| + K = \frac{1}{2}\log\left|\frac{x^2-1}{x^2}\right| + K$$

Applying condition $y=0$ when $x=2$:
$$0 = \frac{1}{2}\log\left|\frac{3}{4}\right| + K \Rightarrow K = -\frac{1}{2}\log\frac{3}{4} = \frac{1}{2}\log\frac{4}{3}$$

Particular solution:
$$y = \frac{1}{2}\log\left|\frac{x^2-1}{x^2}\right| + \frac{1}{2}\log\frac{4}{3}$$

$$\boxed{y = \frac{1}{2}\log\left|\frac{4(x^2-1)}{3x^2}\right|}$$

Given: $\cos\left(\dfrac{dy}{dx}\right) = a$

$$\frac{dy}{dx} = \cos^{-1}a$$

Since $a$ is a constant, $\cos^{-1}a$ is also a constant. Let $\cos^{-1}a = k$.

Separating variables:
$$dy = k\, dx$$

Integrating:
$$y = kx + C = x\cos^{-1}a + C$$

Applying condition $y=1$ when $x=0$:
$$1 = 0 + C \Rightarrow C = 1$$

Particular solution:
$$\boxed{y = x\cos^{-1}a + 1}$$

Given: $\dfrac{dy}{dx} = y\tan x$

Separating variables:
$$\frac{dy}{y} = \tan x\, dx$$

Integrating both sides:
$$\log|y| = \log|\sec x| + \log C$$

$$|y| = C|\sec x|$$

$$y = C\sec x$$

Applying condition $y=1$ when $x=0$:
$$1 = C\sec 0 = C \cdot 1 \Rightarrow C = 1$$

Particular solution:
$$\boxed{y = \sec x}$$

Given: $\dfrac{dy}{dx} = e^x\sin x$, passing through $(0,0)$.

Integrating:
$$y = \int e^x\sin x\, dx$$

Using integration by parts twice:
$$I = \int e^x\sin x\, dx$$
$$I = e^x\sin x - \int e^x\cos x\, dx$$
$$I = e^x\sin x - \left[e^x\cos x + \int e^x\sin x\, dx\right]$$
$$I = e^x\sin x - e^x\cos x - I$$
$$2I = e^x(\sin x - \cos x)$$
$$I = \frac{e^x(\sin x - \cos x)}{2}$$

So: $y = \dfrac{e^x(\sin x - \cos x)}{2} + C$

Applying condition $y=0$ when $x=0$:
$$0 = \frac{e^0(0-1)}{2} + C = -\frac{1}{2} + C \Rightarrow C = \frac{1}{2}$$

Equation of the curve:
$$\boxed{y = \frac{e^x(\sin x - \cos x)}{2} + \frac{1}{2} = \frac{e^x(\sin x - \cos x) + 1}{2}}$$

Given: $xy\dfrac{dy}{dx} = (x+2)(y+2)$

Separating variables:
$$\frac{y}{y+2}dy = \frac{x+2}{x}dx$$

$$\left(1 - \frac{2}{y+2}\right)dy = \left(1 + \frac{2}{x}\right)dx$$

Integrating both sides:
$$y - 2\log|y+2| = x + 2\log|x| + C$$

Applying condition $(1,-1)$:
$$-1 - 2\log|1| = 1 + 2\log|1| + C$$
$$-1 - 0 = 1 + 0 + C \Rightarrow C = -2$$

Equation of the curve:
$$y - 2\log|y+2| = x + 2\log|x| - 2$$

$$\boxed{y - x + 2 = 2\log|x(y+2)|}$$

Given condition: $y\dfrac{dy}{dx} = x$, passing through $(0,-2)$.

Separating variables:
$$y\, dy = x\, dx$$

Integrating both sides:
$$\frac{y^2}{2} = \frac{x^2}{2} + C$$