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Find general solution of dy/dx+2y/x=2e^(x²)xx(x²+1)/x?if y=1 when x=1,express y in terms of x?

Cesareo R.
Featured 1 month ago

See below.

Explanation:

This differential equation is linear so it can be represented as the sum

$y = {y}_{h} + {y}_{p}$ with

$y {'}_{h} + \frac{2}{x} {y}_{h} = 0$
$y {'}_{p} + \frac{2}{x} {y}_{p} = 2 {e}^{{x}^{2}} \left(\frac{{x}^{2} + 1}{x}\right)$

The solution for ${y}_{h}$ is easily determined as

${y}_{h} = {C}_{0} / {x}^{2}$

now supposing that ${y}_{p} = \frac{C \left(x\right)}{x} ^ 2$ and substituting into

$y {'}_{p} + \frac{2}{x} {y}_{p} = 2 {e}^{{x}^{2}} \left(\frac{{x}^{2} + 1}{x}\right)$

we get

$\frac{2 {e}^{{x}^{2}} \left(1 + {x}^{2}\right)}{x} - \frac{C ' \left(x\right)}{x} ^ 2 = 0$ or

$C ' \left(x\right) = {e}^{{x}^{2}} x \left(1 + {x}^{2}\right)$ or

$C \left(x\right) = {e}^{{x}^{2}} {x}^{2} + {C}_{1}$ and finally

$y = \frac{{e}^{{x}^{2}} {x}^{2} + {C}_{1}}{x} ^ 2 = {C}_{1} / {x}^{2} + {e}^{{x}^{2}}$

Differentiate f(x) = lnabs(x/(1+x^2))?

Rhys
Featured 3 weeks ago

$f ' \left(x\right) = \frac{1}{x} - \frac{2 x}{1 + {x}^{2}}$

Explanation:

We must use our log laws:

${\log}_{\alpha} \left(\frac{\beta}{\gamma}\right) \equiv {\log}_{\alpha} \beta - {\log}_{\alpha} \gamma$

$\implies \ln | \frac{x}{1 + {x}^{2}} | = \ln | x | - \ln | 1 + {x}^{2} |$

Now applying our knowledge of differentiating logs:

$\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$

By using the chain rule...

$\frac{d}{\mathrm{dx}} \left(\ln | x | - \ln | 1 + {x}^{2} |\right) = \frac{\frac{d}{\mathrm{dx}} \left(x\right)}{x} - \frac{\frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right)}{1 + {x}^{2}}$

color(blue)(= 1/x - (2x)/(1+x^2)

Let I=int_0^1 9/(3+x^2)^2\ dx. Using the substitution x=sqrt(3) tan(theta), show that I= sqrt(3) int_0^(pi/6) cos^2(theta)\ d theta. What is the exact value of I?

NickTheTurtle
Featured 3 weeks ago

${\int}_{0}^{1} \frac{9}{3 + {x}^{2}} ^ 2 \setminus \mathrm{dx} = \frac{\left(2 \pi + 3\right) \sqrt{3}}{24}$

Explanation:

$\setminus \setminus \setminus \setminus \setminus \setminus {\int}_{0}^{1} \frac{9}{3 + {x}^{2}} ^ 2 \setminus \mathrm{dx}$

Substitute $x = \sqrt{3} \tan \left(\theta\right)$ and $\mathrm{dx} = \sqrt{3} {\sec}^{2} \left(\theta\right) d \setminus \theta$:
$= {\int}_{\arctan} {\left(0\right)}^{\arctan} \left(\frac{1}{\sqrt{3}}\right) \frac{9}{3 + {\left(\sqrt{3} \tan \left(\theta\right)\right)}^{2}} ^ 2 \sqrt{3} {\sec}^{2} \left(\theta\right) \setminus d \setminus \theta$

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} \frac{9 {\sec}^{2} \left(\theta\right)}{3 + 3 {\tan}^{2} \left(\theta\right)} ^ 2 \setminus d \setminus \theta$

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} {\sec}^{2} \frac{\theta}{1 + {\tan}^{2} \left(\theta\right)} ^ 2 \setminus d \setminus \theta$

Using the fact that ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$:

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} {\sec}^{2} \frac{\theta}{\sec} ^ 4 \left(\theta\right) \setminus d \setminus \theta$

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} \frac{1}{\sec} ^ 2 \left(\theta\right) \setminus d \setminus \theta$

Since $\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$, we have

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} {\cos}^{2} \left(\theta\right) \setminus d \setminus \theta$

Using the fact that ${\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$:

$= \frac{\sqrt{3}}{2} {\int}_{0}^{\frac{\pi}{6}} 1 + \cos \left(2 \theta\right) \setminus d \setminus \theta$

$= \frac{\sqrt{3}}{2} {\left[\theta + \sin \frac{2 \theta}{2}\right]}_{0}^{\frac{\pi}{6}}$

$= \frac{\sqrt{3}}{2} \left(\frac{\pi}{6} + \frac{1}{4}\right)$

$= \frac{\left(2 \pi + 3\right) \sqrt{3}}{24}$

A curve is such that dy/dx=4/sqrt((6-2x)) and P(1,8) is a point on the curve. 1. The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR. 2. Find the equation of the curve?

Featured 3 weeks ago

The coordinates of the midpoint of $Q R$ is $= \left(\frac{17}{2} , \frac{17}{4}\right)$. The equation of the curve is y=-4(6-2x))^(1/2)+16

Explanation:

The gradient to the curve $f \left(x\right)$ is

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = \frac{4}{\sqrt{6 - 2 x}}$

At the point $P \left(1 , 8\right)$, the gradient is

$f ' \left(1\right) = \frac{4}{\sqrt{6 - 2 \cdot 1}} = \frac{4}{\sqrt{4}} = 2$

The slope of the tangent at the point $P$ is $m = 2$

Therefore,

The slope of the normal at the point $P$

$= m ' = - \frac{1}{m} = - \frac{1}{2}$

The equation of the normal at the point $P$ is

$y - 8 = - \frac{1}{2} \left(x - 1\right)$

$2 y - 16 = - x + 1$

$2 y + x = 17$

When $x = 0$, $\implies$, $y = \frac{17}{2}$

The point $Q = \left(0 , \frac{17}{2}\right)$

When $y = 0$, $\implies$, $x = 17$

The point $R = \left(17 , 0\right)$

The midpoint of $Q R$ is

$= \left(\frac{17 + 0}{2} , \frac{\frac{17}{2} + 0}{2}\right)$

$= \left(\frac{17}{2} , \frac{17}{4}\right)$

The equation of the curve is obtained by integrating the derivative, or by solving the differential equation

$y = \int \frac{4 \mathrm{dx}}{\sqrt{6 - 2 x}}$

$= 4 \int {\left(6 - 2 x\right)}^{- \frac{1}{2}}$

$= 4 \cdot - \frac{1}{2} \cdot \frac{2}{1} \cdot {\left(6 - 2 x\right)}^{\frac{1}{2}} + C$

y=-4(6-2x))^(1/2)+C

Plugging in the values of $P \left(1 , 8\right)$

$8 = - 4 {\left(6 - 2\right)}^{\frac{1}{2}} + C$

$C = 16$

The equation of the curve is

y=-4(6-2x))^(1/2)+16

graph{(2y+x-17)(y+4sqrt(6-2x)-16)=0 [-22.17, 23.44, -4.57, 18.27]}

Limit when x tends to 0?

Andrea S.
Featured 1 week ago

${\lim}_{x \to 0} {\left({e}^{x} + 3 x\right)}^{\frac{1}{x}} = {e}^{4}$

Explanation:

Write the function as:

${\left({e}^{x} + 3 x\right)}^{\frac{1}{x}} = {\left({e}^{\ln \left({e}^{x} + 3 x\right)}\right)}^{\frac{1}{x}} = {e}^{\ln \frac{{e}^{x} + 3 x}{x}}$

Consider now the limit:

${\lim}_{x \to 0} \ln \frac{{e}^{x} + 3 x}{x}$

It is in the indeterminate form $\frac{0}{0}$ so we can use l'Hospital's rule:

${\lim}_{x \to 0} \ln \frac{{e}^{x} + 3 x}{x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \ln \left({e}^{x} + 3 x\right)}{\frac{d}{\mathrm{dx}} x}$

${\lim}_{x \to 0} \ln \frac{{e}^{x} + 3 x}{x} = {\lim}_{x \to 0} \frac{{e}^{x} + 3}{{e}^{x} + 3 x} = 4$

As the limit is finite and the function ${e}^{x}$ is continuous for $x \in \mathbb{R}$ we have:

${\lim}_{x \to 0} {e}^{\ln \frac{{e}^{x} + 3 x}{x}} = {e}^{\left({\lim}_{x \to 0} \ln \frac{{e}^{x} + 3 x}{x}\right)} = {e}^{4}$

What is the significance of partial derivative? Give an example and help me to understand in brief.

Cesareo R.
Featured yesterday

See below.

Explanation:

I hope it helps.

The partial derivative is intrinsically associated to the total variation.

Suppose we have a function $f \left(x , y\right)$ and we want to know how much it varies when we introduce an increment to each variable.

Fixing ideas, making $f \left(x , y\right) = k x y$ we want to know how much it is

$\mathrm{df} \left(x , y\right) = f \left(x + \mathrm{dx} , y + \mathrm{dy}\right) - f \left(x , y\right)$

In our function-example we have

$f \left(x + \mathrm{dx} , y + \mathrm{dy}\right) = k \left(x + \mathrm{dx}\right) \left(y + \mathrm{dy}\right) = k x y + k x \mathrm{dx} + k y \mathrm{dy} + k \mathrm{dx} \mathrm{dy}$

and then

$\mathrm{df} \left(x , y\right) = k x y + k x \mathrm{dx} + k y \mathrm{dy} + k \mathrm{dx} \mathrm{dy} - k x y = k x \mathrm{dx} + k y \mathrm{dy} + k \mathrm{dx} \mathrm{dy}$

Choosing $\mathrm{dx} , \mathrm{dy}$ arbitrarily small then $\mathrm{dx} \mathrm{dy} \approx 0$ and then

$\mathrm{df} \left(x , y\right) = k x \mathrm{dx} + k y \mathrm{dy}$

but generally

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

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

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

now making $\mathrm{dx} , \mathrm{dy}$ arbitrarily small we have

$\mathrm{df} \left(x , y\right) = \frac{1}{2} \left(2 {f}_{x} \left(x , y\right) \mathrm{dx} + 2 {f}_{y} \left(x , y\right) \mathrm{dy}\right) = {f}_{x} \left(x , y\right) \mathrm{dx} + {f}_{y} \left(x , y\right) \mathrm{dy}$

so we can compute the total variation for a given function, by calculating the partial derivatives ${f}_{{x}_{1}} , {f}_{{x}_{2}} , \cdots , {f}_{{x}_{n}}$ and compounding

$\mathrm{df} \left({x}_{1} , {x}_{2} , \cdots , {x}_{n}\right) = {f}_{{x}_{1}} {\mathrm{dx}}_{1} + \cdots + {f}_{{x}_{n}} {\mathrm{dx}}_{n}$

Here, the quantities ${f}_{{x}_{i}}$ are called partial derivatives and can also be represented as

$\frac{\partial f}{\partial {x}_{i}}$

In our example

${f}_{x} = \frac{\partial f}{\partial x} = k x$ and

${f}_{y} = \frac{\partial f}{\partial y} = k y$

NOTE

${f}_{x} \left(x , y\right) = {\lim}_{\begin{matrix}\mathrm{dx} \to 0 \\ \mathrm{dy} \to 0\end{matrix}} \frac{f \left(x + \mathrm{dx} , y\right) - f \left(x , y\right)}{\mathrm{dx}} = {\lim}_{\begin{matrix}\mathrm{dx} \to 0 \\ \mathrm{dy} \to 0\end{matrix}} \frac{f \left(x + \mathrm{dx} , y + \mathrm{dy}\right) - f \left(x , y\right)}{\mathrm{dx}}$

${f}_{y} \left(x , y\right) = {\lim}_{\begin{matrix}\mathrm{dx} \to 0 \\ \mathrm{dy} \to 0\end{matrix}} \frac{f \left(x , y + \mathrm{dy}\right) - f \left(x , y\right)}{\mathrm{dy}} = {\lim}_{\begin{matrix}\mathrm{dx} \to 0 \\ \mathrm{dy} \to 0\end{matrix}} \frac{f \left(x + \mathrm{dx} , y + \mathrm{dy}\right) - f \left(x , y\right)}{\mathrm{dy}}$

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