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\((1, -5, -6)\)

We are given two points, \(A = (3, 6, -1)\) and \(B = (6, 2, -2)\). The point \(P\) lies on the line segment \(AB\), and its \(y\)-coordinate is \(4\). We need to find its \(z\)-coordinate.

Let \(P\) divide the line segment \(AB\) in the ratio \(\lambda:1\).

Step 1: Find the Ratio (\(\lambda\))

We use the section formula for the \(y\)-coordinate, where \(y=4\), \(y_1=6\), and \(y_2=2\):

The point \(P\) is the **midpoint** of the segment \(AB\) since \(\lambda = 1\).

Step 2: Find the \(z\)-coordinate (\(z\))

Now, we use the section formula for the \(z\)-coordinate with \(\lambda=1\), \(z_1=-1\), and \(z_2=-2\):

A function $f(x, y)$ is said to be homogeneous of degree $k$ if $f(tx, ty) = t^k f(x, y)$ for some real number $k$ and all $t > 0$. To identify a function that is not homogeneous, we look for functions where this scaling property does not hold. Common characteristics of non-homogeneous functions include: 1. **Presence of a constant term:** A term that does not involve $x$ or $y$. 2. **Terms with different total degrees:** For polynomial functions, if the sum of the powers of $x$ and $y$ in each term is not the same. 3. **Transcendental functions (e.g., exponential, logarithmic) that do not scale appropriately:** For instance, $e^x$, $\log(x+y)$, or $\sin(x)$ are generally not homogeneous unless their arguments are homogeneous of degree 0. Let's consider a common example of a function that is not homogeneous: The function $f(x, y) = x^2 + y + 1$ is not a homogeneous function. **Step-by-step verification:** 1. **Recall the definition of a homogeneous function:** $f(tx, ty) = t^k f(x, y)$. 2. **Substitute $tx$ for $x$ and $ty$ for $y$ into the function $f(x, y) = x^2 + y + 1$:** $$f(tx, ty) = (tx)^2 + (ty) + 1$$ 3. **Simplify the expression:** $$f(tx, ty) = t^2x^2 + ty + 1$$ 4. **Compare $f(tx, ty)$ with $t^k f(x, y)$:** If $f(x, y)$ were homogeneous, we should be able to factor out $t^k$ such that the remaining expression is $x^2 + y + 1$. However, we cannot write $t^2x^2 + ty + 1$ as $t^k(x^2 + y + 1)$ for any constant $k$. For instance, the term $1$ does not scale with $t$, and the term $ty$ scales differently from $t^2x^2$. Since $f(tx, ty) \ne t^k f(x, y)$ for any real number $k$, the function $f(x, y) = x^2 + y + 1$ is not a homogeneous function.

The given differential equation is: \[ (x + 2y^3)\frac{dy}{dx} = 2y \] We rearrange the equation to the linear form $\frac{dx}{dy} + P(y)x = Q(y)$: \[ \frac{dx}{dy} = \frac{x + 2y^3}{2y} = \frac{x}{2y} + \frac{2y^3}{2y} = \frac{1}{2y}x + y^2 \] \[ \frac{dx}{dy} - \frac{1}{2y}x = y^2 \] This is a linear differential equation in $x$ with $P(y) = -\frac{1}{2y}$. The integrating factor (IF) is given by: \[ IF = e^{\int P(y) dy} \] Substitute $P(y)$: \[ IF = e^{\int \left(-\frac{1}{2y}\right) dy} = e^{-\frac{1}{2} \int \frac{1}{y} dy} \] \[ IF = e^{-\frac{1}{2} \ln |y|} \] Using the logarithm property $a \ln b = \ln b^a$: \[ IF = e^{\ln |y|^{-\frac{1}{2}}} = |y|^{-\frac{1}{2}} \] Assuming $y > 0$, the integrating factor is: \[ IF = \frac{1}{\sqrt{y}} \]

The given differential equation is:

$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)^3 = 0$

Step 1: Expand the derivative

$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)^3 = 3\left(\frac{dy}{dx}\right)^2 \frac{d^2y}{dx^2} = 0$

Step 2: Identify order and degree

• The highest order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $p = 2$.
• The equation is a polynomial in derivatives, and the degree of the highest derivative is $1$, so $q = 1$.

Step 3: Compute $(p - q)$

$(p - q) = 2 - 1 = 1$

∴ Final Answer: $(p - q) = 1$

The order of a differential equation is the order of the highest derivative present in the equation.The degree of a differential equation is the power of the highest order derivative, provided the differential equation is a polynomial equation in its derivatives.

For the degree to be defined, the equation must be free from terms like $\sin(\frac{dy}{dx})$, $e^{\frac{dy}{dx}}$, $\log(\frac{d^2y}{dx^2})$, etc. (i.e., the derivatives must not appear inside transcendental functions).In the given equation, the term $x\sin\left(\frac{dy}{dx}\right)$ involves the first derivative $\frac{dy}{dx}$ inside the transcendental function $\sin()$.Since the equation is not a polynomial in its derivatives, the Degree is not defined