Differential Equations and Modeling — NCERT Solutions

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

Concept: The order of a differential equation is the order of the highest-order derivative present. The degree is the power of that highest-order derivative after the equation is made free of radicals and fractions in derivatives.

Working:

• The highest-order derivative present is $\dfrac{dy}{dx}$ (first derivative).
• It appears with power 1.

Answer: Order = 1, Degree = 1

Given: $\dfrac{dy}{dx} + e^y = 0$

Concept: Order = order of highest derivative; Degree = power of highest derivative (must be a polynomial in derivatives).

Working:

• The highest-order derivative is $\dfrac{dy}{dx}$ (first derivative), appearing with power 1.
• The term $e^y$ involves $y$ (not a derivative), so it does not affect the degree.

Answer: Order = 1, Degree = 1

Given: $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$

Working:

• The highest-order derivative is $\dfrac{d^2y}{dx^2}$ (second derivative).
• It appears with power 1.

Answer: Order = 2, Degree = 1

Given: $\left(\dfrac{dy}{dx}\right)^4 + 3y\left(\dfrac{d^3y}{dx^3}\right) = 0$

Working:

• Derivatives present: $\dfrac{dy}{dx}$ (order 1) and $\dfrac{d^3y}{dx^3}$ (order 3).
• Highest-order derivative: $\dfrac{d^3y}{dx^3}$ — this is order 3.
• The highest-order derivative $\dfrac{d^3y}{dx^3}$ appears with power 1.

Answer: Order = 3, Degree = 1

(Note: The answer key states order 2, degree 1, but based on the equation as written the highest derivative is third order. If the second term were $\dfrac{d^2y}{dx^2}$, the answer would be order 2, degree 1. Students should follow the equation as printed.)

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

As stated in the problem: $y' = \dfrac{dy}{dx}$, $y'' = \dfrac{d^2y}{dx^2}$, $y''' = \dfrac{d^3y}{dx^3}$.

Working:

• Derivatives present: $y'$ (order 1), $y''$ (order 2), $y'''$ (order 3).
• Highest-order derivative: $y'''$ — order = 3.
• The highest-order derivative $y'''$ appears as $(y''')^2$, so its power = 2.
• The equation is already a polynomial in derivatives.

Answer: Order = 3, Degree = 2

Given: $y = ae^{-x}$

Step 1 – Differentiate:
$$\frac{dy}{dx} = -ae^{-x}$$

Step 2 – Substitute into the LHS of the DE:
$$\frac{dy}{dx} + y = -ae^{-x} + ae^{-x} = 0 = \text{RHS}$$

Conclusion: Since LHS = RHS, $y = ae^{-x}$ is a solution of $\dfrac{dy}{dx} + y = 0$. ✓

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

Step 1 – Differentiate:
$$\frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$$

Step 2 – Compute RHS:
$$\frac{xy}{1+x^2} = \frac{x \cdot \sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$$

Step 3 – Compare:
$$\text{LHS} = \frac{x}{\sqrt{1+x^2}} = \text{RHS}$$

Conclusion: $y = \sqrt{1+x^2}$ is a solution. ✓

Given: $xy = \log y + c$

Step 1 – Differentiate implicitly with respect to $x$:
$$y + x\frac{dy}{dx} = \frac{1}{y}\frac{dy}{dx}$$

Step 2 – Collect $\dfrac{dy}{dx}$ terms:
$$y = \frac{dy}{dx}\left(\frac{1}{y} - x\right) = \frac{dy}{dx}\cdot\frac{1-xy}{y}$$

Step 3 – Solve for $\dfrac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{y \cdot y}{1-xy} = \frac{y^2}{1-xy}$$

Conclusion: This equals the RHS of the given DE. Hence $xy = \log y + c$ is a solution. ✓

Given: $ax^2 + by^2 = 1$ … (i)

Step 1 – First differentiation (w.r.t. $x$):
$$2ax + 2by\,y_1 = 0 \implies ax + by\,y_1 = 0 \quad \dots (ii)$$

Step 2 – Second differentiation:
$$a + b(y_1^2 + y\,y_2) = 0 \quad \dots (iii)$$

Step 3 – Eliminate $a$ and $b$.

From (ii): $a = -\dfrac{by\,y_1}{x}$

Substitute into (iii):
$$-\frac{by\,y_1}{x} + b(y_1^2 + y\,y_2) = 0$$

Divide by $b$ (assuming $b \neq 0$):
$$-\frac{y\,y_1}{x} + y_1^2 + y\,y_2 = 0$$

Multiply throughout by $x$:
$$-y\,y_1 + x\,y_1^2 + xy\,y_2 = 0$$
$$\Rightarrow x(y\,y_2 + y_1^2) = y\,y_1$$

Conclusion: This is exactly the given DE. Hence $ax^2 + by^2 = 1$ is a solution. ✓

Given: $y = (a+bx)e^{2x}$

Step 1 – First derivative:
$$y_1 = b\,e^{2x} + 2(a+bx)e^{2x} = e^{2x}[b + 2(a+bx)] = e^{2x}(2a+b+2bx)$$

Step 2 – Second derivative:
$$y_2 = 2e^{2x}(2a+b+2bx) + e^{2x}(2b) = e^{2x}(4a+2b+4bx+2b) = e^{2x}(4a+4b+4bx)$$

Step 3 – Substitute into $y_2 - 4y_1 + 4y$:
$$y_2 - 4y_1 + 4y = e^{2x}(4a+4b+4bx) - 4e^{2x}(2a+b+2bx) + 4e^{2x}(a+bx)$$
$$= e^{2x}\left[(4a+4b+4bx) - (8a+4b+8bx) + (4a+4bx)\right]$$
$$= e^{2x}\left[4a+4b+4bx - 8a-4b-8bx+4a+4bx\right]$$
$$= e^{2x}\left[(4a-8a+4a)+(4b-4b)+(4bx-8bx+4bx)\right]$$
$$= e^{2x}[0+0+0] = 0$$

Conclusion: $y = (a+bx)e^{2x}$ is a solution of $y_2 - 4y_1 + 4y = 0$. ✓

Given: $x^2 = 2y^2 \log y$ … (i)

Step 1 – Differentiate implicitly w.r.t. $x$:
$$2x = 2\left[2y\log y \cdot \frac{dy}{dx} + y^2 \cdot \frac{1}{y}\cdot\frac{dy}{dx}\right]$$
$$2x = 2\frac{dy}{dx}\left[2y\log y + y\right]$$
$$x = \frac{dy}{dx}\cdot y(2\log y + 1)$$

$$\Rightarrow \frac{dy}{dx} = \frac{x}{y(2\log y+1)} \quad \dots (ii)$$

Step 2 – From (i): $2\log y = \dfrac{x^2}{y^2}$, so $2\log y + 1 = \dfrac{x^2+y^2}{y^2}$.

Step 3 – Substitute into (ii):
$$\frac{dy}{dx} = \frac{x}{y \cdot \dfrac{x^2+y^2}{y^2}} = \frac{xy}{x^2+y^2}$$

Step 4 – Check the DE:
$$(x^2+y^2)\frac{dy}{dx} - xy = (x^2+y^2)\cdot\frac{xy}{x^2+y^2} - xy = xy - xy = 0$$

Conclusion: $x^2 = 2y^2\log y$ is a solution. ✓

Given: $y = ke^x - 1$

Part 1 – Verification:

Differentiate: $\dfrac{dy}{dx} = ke^x$

RHS of DE: $y + 1 = ke^x - 1 + 1 = ke^x$

Since LHS = RHS, $y = ke^x - 1$ is a solution. ✓

Part 2 – Finding $k$:

The curve passes through $(0, 1)$, so substitute $x = 0$, $y = 1$:
$$1 = ke^0 - 1 = k - 1$$
$$\Rightarrow k = 2$$

Answer: $k = \boxed{2}$

Given: $x^2 - y^2 = a^2$ … (i)

Step 1 – Differentiate w.r.t. $x$:
$$2x - 2y\frac{dy}{dx} = 0$$
$$\Rightarrow x - y\,y_1 = 0$$
$$\Rightarrow y\,y_1 = x$$

Answer: The required differential equation is $\boxed{y\,y_1 = x}$ (i.e., $y\dfrac{dy}{dx} = x$).

Given: Family of circles with centre at origin: $x^2 + y^2 = r^2$, where $r$ is an arbitrary constant.

Step 1 – Differentiate w.r.t. $x$:
$$2x + 2y\frac{dy}{dx} = 0$$
$$\Rightarrow x + y\,y_1 = 0$$

Answer: The required differential equation is $\boxed{x + y\dfrac{dy}{dx} = 0}$.

Given: A circle with centre on the $y$-axis has centre $(0, k)$ for some constant $k$. Since it passes through the origin $(0,0)$, its radius $= k$.

Equation: $x^2 + (y-k)^2 = k^2$
$$\Rightarrow x^2 + y^2 - 2ky = 0 \quad \dots (i)$$

Step 1 – Differentiate w.r.t. $x$:
$$2x + 2y\frac{dy}{dx} - 2k\frac{dy}{dx} = 0$$
$$\Rightarrow 2x + 2(y-k)y_1 = 0 \quad \dots (ii)$$

Step 2 – Eliminate $k$.

From (i): $k = \dfrac{x^2+y^2}{2y}$

Substitute into (ii):
$$2x + 2\left(y - \frac{x^2+y^2}{2y}\right)y_1 = 0$$
$$2x + 2\cdot\frac{2y^2 - x^2 - y^2}{2y}\cdot y_1 = 0$$
$$2x + \frac{(y^2-x^2)}{y}\cdot y_1 = 0$$

Multiply by $y$:
$$2xy + (y^2 - x^2)y_1 = 0$$

Answer: $\boxed{(y^2 - x^2)\dfrac{dy}{dx} + 2xy = 0}$

Given: $y = e^{2x}(a+bx)$ … (i)

Since there are two arbitrary constants, we differentiate twice.

Step 1 – First derivative:
$$y_1 = 2e^{2x}(a+bx) + be^{2x} = e^{2x}(2a+2bx+b)$$
$$\Rightarrow y_1 = 2y + be^{2x} \quad \dots (ii)$$

(using $y = e^{2x}(a+bx)$)

Step 2 – Second derivative:
$$y_2 = 2y_1 + 2be^{2x} \quad \dots (iii)$$

Step 3 – Eliminate $b$.

From (ii): $be^{2x} = y_1 - 2y$

Substitute into (iii):
$$y_2 = 2y_1 + 2(y_1 - 2y)$$
$$y_2 = 2y_1 + 2y_1 - 4y$$
$$y_2 = 4y_1 - 4y$$

Answer: $\boxed{y_2 - 4y_1 + 4y = 0}$, i.e., $\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 4y = 0$.

Given: Parabolas with vertex at origin and focus on positive $x$-axis have the form:
$$y^2 = 4ax, \quad a > 0 \quad \dots (i)$$

Step 1 – Differentiate w.r.t. $x$:
$$2y\frac{dy}{dx} = 4a \implies 4a = 2y\,y_1 \quad \dots (ii)$$

Step 2 – Substitute (ii) into (i) to eliminate $a$:
$$y^2 = (2y\,y_1)\cdot x$$
$$y^2 = 2xy\,y_1$$
$$y = 2x\,y_1$$

Answer: $\boxed{y = 2x\dfrac{dy}{dx}}$

Given: Standard ellipse with centre at origin and foci on $x$-axis:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0 \quad \dots (i)$$

Two arbitrary constants $a$ and $b$ ⟹ differentiate twice.

Step 1 – Differentiate (i) w.r.t. $x$:
$$\frac{2x}{a^2} + \frac{2y\,y_1}{b^2} = 0$$
$$\Rightarrow \frac{x}{a^2} + \frac{y\,y_1}{b^2} = 0 \quad \dots (ii)$$

Step 2 – Differentiate (ii) w.r.t. $x$:
$$\frac{1}{a^2} + \frac{y_1^2 + y\,y_2}{b^2} = 0 \quad \dots (iii)$$

Step 3 – Eliminate $a^2$ and $b^2$.

From (ii): $\dfrac{1}{a^2} = -\dfrac{y\,y_1}{b^2 x}$

Substitute into (iii):
$$-\frac{y\,y_1}{b^2 x} + \frac{y_1^2 + y\,y_2}{b^2} = 0$$

Multiply by $b^2$:
$$-\frac{y\,y_1}{x} + y_1^2 + y\,y_2 = 0$$

Multiply by $x$:
$$-y\,y_1 + x\,y_1^2 + xy\,y_2 = 0$$
$$\Rightarrow xy\,y_2 + x\,y_1^2 - y\,y_1 = 0$$

Answer: $\boxed{xy\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 - y\dfrac{dy}{dx} = 0}$

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

Method: Separation of variables.

Step 1 – Separate variables:
$$\frac{dy}{y} = (e^x+1)\,dx$$

Step 2 – Integrate both sides:
$$\int \frac{dy}{y} = \int (e^x+1)\,dx$$
$$\ln|y| = e^x + x + C$$

Answer: $\boxed{\ln|y| = e^x + x + C}$, or equivalently $y = Ae^{e^x+x}$ where $A = e^C$.

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

Step 1 – Separate variables:
$$\frac{dy}{y^5} = -\frac{dx}{x^5}$$
$$y^{-5}\,dy = -x^{-5}\,dx$$

Step 2 – Integrate both sides:
$$\int y^{-5}\,dy = -\int x^{-5}\,dx$$
$$\frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C$$
$$-\frac{1}{4y^4} = \frac{1}{4x^4} + C$$

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

Answer: $\boxed{\dfrac{1}{x^4} + \dfrac{1}{y^4} = C}$ (where $C$ is an arbitrary constant).