Differential Equations worksheet with 18 problems covering order, degree, and solving techniques.

Problem: Solve the given differential equations and explain the solution.

#### Problem 1: Find the order and degree of the following differential equations.

1. Equation: $\frac{d^3y}{dx^3} + \cos\left(\frac{d^2y}{dx^2}\right) = 3$

- Order: The highest derivative present in the equation is $\frac{d^3y}{dx^3}$, which is a third-order derivative. Therefore, the order of the differential equation is 3.
- Degree: The degree of a differential equation is defined as the power of the highest-order derivative when the equation is expressed as a polynomial in derivatives. However, in this case, the term $\cos\left(\frac{d^2y}{dx^2}\right)$ is not a polynomial in derivatives. Hence, the degree is not defined.

Answer: Order = 3, Degree = Not Defined

2. Equation: $\frac{d^4y}{dx^4} + \cos(y'') = 3$

- Order: The highest derivative present in the equation is $\frac{d^4y}{dx^4}$, which is a fourth-order derivative. Therefore, the order of the differential equation is 4.
- Degree: The term $\cos(y'')$ is not a polynomial in derivatives. Hence, the degree is not defined.

Answer: Order = 4, Degree = Not Defined

3. Equation: $(y'')^2 + (y')^2 + y^4 = 0$

- Order: The highest derivative present in the equation is $y''$, which is a second-order derivative. Therefore, the order of the differential equation is 2.
- Degree: The equation is a polynomial in derivatives, and the highest power of the highest-order derivative ($y''$) is 2. Therefore, the degree is 2.

Answer: Order = 2, Degree = 2

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#### Problem 2: Form the differential equation representing the family of ellipses having foci on the x-axis and the center at the origin.

The general equation of an ellipse with foci on the x-axis and center at the origin is:
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\]
where $a > b > 0$.

To eliminate the constants $a$ and $b$, we differentiate the equation with respect to $x$:
\[
\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0
\]
Simplifying, we get:
\[
\frac{x}{a^2} + \frac{y}{b^2} \frac{dy}{dx} = 0 \quad \text{(1)}
\]

Differentiate again with respect to $x$:
\[
\frac{1}{a^2} + \frac{1}{b^2} \left( \frac{dy}{dx} \right)^2 + \frac{y}{b^2} \frac{d^2y}{dx^2} = 0
\]
Simplifying, we get:
\[
\frac{1}{a^2} + \frac{\left( \frac{dy}{dx} \right)^2}{b^2} + \frac{y}{b^2} \frac{d^2y}{dx^2} = 0 \quad \text{(2)}
\]

From equation (1), solve for $\frac{1}{a^2}$:
\[
\frac{1}{a^2} = -\frac{y}{b^2} \frac{dy}{dx}
\]

Substitute $\frac{1}{a^2}$ into equation (2):
\[
-\frac{y}{b^2} \frac{dy}{dx} + \frac{\left( \frac{dy}{dx} \right)^2}{b^2} + \frac{y}{b^2} \frac{d^2y}{dx^2} = 0
\]
Multiply through by $b^2$:
\[
-y \frac{dy}{dx} + \left( \frac{dy}{dx} \right)^2 + y \frac{d^2y}{dx^2} = 0
\]
Rearrange the terms:
\[
y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0
\]

Thus, the differential equation is:
\[
\boxed{y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0}
\]

---

#### Problem 3: Form the differential equation of the family of circles in the quadrant which touch the coordinate axes.

The general equation of a circle touching both the x-axis and y-axis in the first quadrant is:
\[
(x - r)^2 + (y - r)^2 = r^2
\]
where $r$ is the radius of the circle.

Expand the equation:
\[
x^2 - 2rx + r^2 + y^2 - 2ry + r^2 = r^2
\]
Simplify:
\[
x^2 + y^2 - 2rx - 2ry + r^2 = 0
\]

To eliminate the constant $r$, differentiate with respect to $x$:
\[
2x - 2r - 2y \frac{dy}{dx} - 2r \frac{dy}{dx} = 0
\]
Simplify:
\[
x - r - y \frac{dy}{dx} - r \frac{dy}{dx} = 0
\]
Rearrange to solve for $r$:
\[
r(1 + \frac{dy}{dx}) = x - y \frac{dy}{dx}
\]
\[
r = \frac{x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}}
\]

Substitute $r$ back into the original equation $x^2 + y^2 - 2rx - 2ry + r^2 = 0$:
\[
x^2 + y^2 - 2 \left( \frac{x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} \right) x - 2 \left( \frac{x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} \right) y + \left( \frac{x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} \right)^2 = 0
\]

After simplification, the differential equation is:
\[
(x + y) \frac{dy}{dx} = x - y
\]

Thus, the differential equation is:
\[
\boxed{(x + y) \frac{dy}{dx} = x - y}
\]

---

#### Problem 4: Solve $xy(y + 1) dy = (x^2 + 1) dx$

Rewrite the equation:
\[
\frac{y(y + 1)}{x^2 + 1} dy = \frac{dx}{x}
\]

Separate variables:
\[
\int \frac{y(y + 1)}{x^2 + 1} dy = \int \frac{dx}{x}
\]

Integrate both sides:
\[
\int \left( \frac{y^2}{x^2 + 1} + \frac{y}{x^2 + 1} \right) dy = \int \frac{dx}{x}
\]

The left-hand side can be split into two integrals:
\[
\int \frac{y^2}{x^2 + 1} dy + \int \frac{y}{x^2 + 1} dy
\]

However, since $x^2 + 1$ is independent of $y$, we can treat it as a constant with respect to $y$. Thus, the integral becomes:
\[
\frac{1}{x^2 + 1} \int (y^2 + y) dy = \ln|x| + C
\]

Integrate $y^2 + y$:
\[
\frac{1}{x^2 + 1} \left( \frac{y^3}{3} + \frac{y^2}{2} \right) = \ln|x| + C
\]

Multiply through by $x^2 + 1$:
\[
\frac{y^3}{3} + \frac{y^2}{2} = (x^2 + 1)(\ln|x| + C)
\]

Thus, the solution is:
\[
\boxed{\frac{y^3}{3} + \frac{y^2}{2} = (x^2 + 1)(\ln|x| + C)}
\]

---

#### **Problem 5: Solve $(1 - y)x \frac{dy}{dx} + (1 + y)x = 0$

Rewrite the equation:
\[
(1 - y)x \frac{dy}{dx} = -(1 + y)x
\]

Separate variables:
\[
\frac{1 - y}{1 + y} dy = -\frac{dx}{x}
\]

Integrate both sides:
\[
\int \frac{1 - y}{1 + y} dy = -\int \frac{dx}{x}
\]

The left-hand side can be simplified using partial fractions:
\[
\frac{1 - y}{1 + y} = \frac{-(y + 1) + 2}{1 + y} = -1 + \frac{2}{1 + y}
\]

Thus, the integral becomes:
\[
\int \left( -1 + \frac{2}{1 + y} \right) dy = -\int \frac{dx}{x}
\]

Integrate:
\[
-\int 1 \, dy + 2 \int \frac{1}{1 + y} dy = -\ln|x| + C
\]
\[
-y + 2 \ln|1 + y| = -\ln|x| + C
\]

Rearrange:
\[
2 \ln|1 + y| - y = -\ln|x| + C
\]

Thus, the solution is:
\[
\boxed{2 \ln|1 + y| - y = -\ln|x| + C}
\]

---

#### Problem 6: Solve $\frac{dy}{dx} = 1 - \frac{\cos x}{1 + \cos x}$

Simplify the right-hand side:
\[
\frac{dy}{dx} = 1 - \frac{\cos x}{1 + \cos x} = \frac{1 + \cos x - \cos x}{1 + \cos x} = \frac{1}{1 + \cos x}
\]

Separate variables:
\[
dy = \frac{dx}{1 + \cos x}
\]

Integrate both sides:
\[
\int dy = \int \frac{dx}{1 + \cos x}
\]

Use the trigonometric identity $1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)$:
\[
\int dy = \int \frac{dx}{2 \cos^2\left(\frac{x}{2}\right)} = \frac{1}{2} \int \sec^2\left(\frac{x}{2}\right) dx
\]

Integrate:
\[
y = \frac{1}{2} \cdot 2 \tan\left(\frac{x}{2}\right) + C = \tan\left(\frac{x}{2}\right) + C
\]

Thus, the solution is:
\[
\boxed{y = \tan\left(\frac{x}{2}\right) + C}
\]

---

#### **Problem 7: Solve $(\sin x + \cos x) dy + (\cos x - \sin x) dx = 0$

Rewrite the equation:
\[
(\sin x + \cos x) dy = -(\cos x - \sin x) dx
\]

Separate variables:
\[
\frac{dy}{\cos x - \sin x} = -\frac{dx}{\sin x + \cos x}
\]

Integrate both sides:
\[
\int \frac{dy}{\cos x - \sin x} = -\int \frac{dx}{\sin x + \cos x}
\]

Let $I_1 = \int \frac{dx}{\sin x + \cos x}$ and $I_2 = \int \frac{dx}{\cos x - \sin x}$. Using the substitution $u = \tan\left(\frac{x}{2}\right)$, we find:
\[
I_1 = \frac{1}{\sqrt{2}} \ln\left| \tan\left(\frac{x}{2} + \frac{\pi}{8}\right) \right| + C_1
\]
\[
I_2 = \frac{1}{\sqrt{2}} \ln\left| \tan\left(\frac{x}{2} - \frac{\pi}{8}\right) \right| + C_2
\]

Thus, the solution is:
\[
\boxed{y = \frac{1}{\sqrt{2}} \ln\left| \frac{\cos x - \sin x}{\sin x + \cos x} \right| + C}
\]

---

#### Problem 8: Solve $\sec^2 y (1 + x^2) dy + 2x \tan y dx = 0$ given that $y(1) = \frac{\pi}{4}$

Rewrite the equation:
\[
\sec^2 y (1 + x^2) dy = -2x \tan y dx
\]

Separate variables:
\[
\frac{\sec^2 y}{\tan y} dy = -\frac{2x}{1 + x^2} dx
\]

Simplify:
\[
\frac{\sec^2 y}{\tan y} = \frac{1}{\cos^2 y} \cdot \frac{\cos y}{\sin y} = \frac{1}{\cos y \sin y} = \frac{2}{\sin 2y}
\]

Thus, the equation becomes:
\[
\frac{2}{\sin 2y} dy = -\frac{2x}{1 + x^2} dx
\]

Integrate both sides:
\[
\int \frac{2}{\sin 2y} dy = -\int \frac{2x}{1 + x^2} dx
\]

The left-hand side is:
\[
\int \frac{2}{\sin 2y} dy = \int \csc 2y \, d(2y) = \ln|\csc 2y - \cot 2y| + C_1
\]

The right-hand side is:
\[
-\int \frac{2x}{1 + x^2} dx = -\ln|1 + x^2| + C_2
\]

Thus:
\[
\ln|\csc 2y - \cot 2y| = -\ln|1 + x^2| + C
\]

Exponentiate both sides:
\[
|\csc 2y - \cot 2y| = \frac{C}{1 + x^2}
\]

Using the initial condition $y(1) = \frac{\pi}{4}$:
\[
\csc\left(\frac{\pi}{2}\right) - \cot\left(\frac{\pi}{2}\right) = 1 - 0 = 1
\]
\[
1 = \frac{C}{1 + 1^2} \implies C = 2
\]

Thus, the solution is:
\[
\boxed{\csc 2y - \cot 2y = \frac{2}{1 + x^2}}
\]

---

#### Problem 9: Solve $(1 + y^2)(1 + \log x) dx + xy dy = 0$ given $y(1) = 1$

Rewrite the equation:
\[
(1 + y^2)(1 + \log x) dx = -xy dy
\]

Separate variables:
\[
\frac{(1 + \log x) dx}{x} = -\frac{y dy}{1 + y^2}
\]

Integrate both sides:
\[
\int \frac{1 + \log x}{x} dx = -\int \frac{y}{1 + y^2} dy
\]

The left-hand side is:
\[
\int \frac{1 + \log x}{x} dx = \int \frac{1}{x} dx + \int \frac{\log x}{x} dx = \ln|x| + \frac{(\log x)^2}{2} + C_1
\]

The right-hand side is:
\[
-\int \frac{y}{1 + y^2} dy = -\frac{1}{2} \ln|1 + y^2| + C_2
\]

Thus:
\[
\ln|x| + \frac{(\log x)^2}{2} = -\frac{1}{2} \ln|1 + y^2| + C
\]

Using the initial condition $y(1) = 1$:
\[
\ln|1| + \frac{(\log 1)^2}{2} = -\frac{1}{2} \ln|1 + 1^2| + C \implies 0 = -\frac{1}{2} \ln 2 + C \implies C = \frac{1}{2} \ln 2
\]

Thus, the solution is:
\[
\boxed{\ln|x| + \frac{(\log x)^2}{2} + \frac{1}{2} \ln|1 + y^2| = \frac{1}{2} \ln 2}
\]

---

#### Problem 10: Solve $\log\left(\frac{dy}{dx}\right) = 3x + 4y$ given that $y = 0$ when $x = 0$

Rewrite the equation:
\[
\frac{dy}{dx} = e^{3x + 4y}
\]

Separate variables:
\[
e^{-4y} dy = e^{3x} dx
\]

Integrate both sides:
\[
\int e^{-4y} dy = \int e^{3x} dx
\]

The left-hand side is:
\[
\int e^{-4y} dy = -\frac{1}{4} e^{-4y} + C_1
\]

The right-hand side is:
\[
\int e^{3x} dx = \frac{1}{3} e^{3x} + C_2
\]

Thus:
\[
-\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} + C
\]

Using the initial condition $y = 0$ when $x = 0$:
\[
-\frac{1}{4} e^0 = \frac{1}{3} e^0 + C \implies -\frac{1}{4} = \frac{1}{3} + C \implies C = -\frac{1}{4} - \frac{1}{3} = -\frac{3}{12} - \frac{4}{12} = -\frac{7}{12}
\]

Thus, the solution is:
\[
\boxed{-\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} - \frac{7}{12}}
\]

---

#### **Problem 11: Solve $x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x$

Let $v = \frac{y}{x} \implies y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substitute into the equation:
\[
x \cos(v) \left( v + x \frac{dv}{dx} \right) = vx \cos(v) + x
\]

Simplify:
\[
x \cos(v) v + x^2 \cos(v) \frac{dv}{dx} = vx \cos(v) + x
\]
\[
x^2 \cos(v) \frac{dv}{dx} = x
\]
\[
\cos(v) \frac{dv}{dx} = \frac{1}{x}
\]

Separate variables:
\[
\cos(v) dv = \frac{dx}{x}
\]

Integrate both sides:
\[
\int \cos(v) dv = \int \frac{dx}{x}
\]

The left-hand side is:
\[
\int \cos(v) dv = \sin(v) + C_1
\]

The right-hand side is:
\[
\int \frac{dx}{x} = \ln|x| + C_2
\]

Thus:
\[
\sin(v) = \ln|x| + C
\]

Substitute back $v = \frac{y}{x}$:
\[
\sin\left(\frac{y}{x}\right) = \ln|x| + C
\]

Thus, the solution is:
\[
\boxed{\sin\left(\frac{y}{x}\right) = \ln|x| + C}
\]

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#### **Problem 12: Solve $\left[x \cos\left(\frac{y}{x}\right) - y \sin\left(\frac{y}{x}\right)\right] dx = \left[y \sin\left(\