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Solve the differential equation $\frac{1}{\left(y-1\right)^2}dx+\frac{1}{\sqrt{x^2+4}}dy=0$

Step-by-step Solution

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Final Answer

$\frac{-\left(y-1\right)^{3}}{3}=4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$
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Step-by-step Solution

Specify the solving method

1

Group the terms of the equation

$\frac{1}{\sqrt{x^2+4}}dy=-\left(\frac{1}{\left(y-1\right)^2}\right)dx$
2

Multiplying the fraction by $-1$

$\frac{1}{\sqrt{x^2+4}}dy=\frac{-1}{\left(y-1\right)^2}dx$
3

Divide fractions $\frac{1}{\frac{-1}{\left(y-1\right)^2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$-\left(y-1\right)^2=\frac{1}{\frac{1}{\sqrt{x^2+4}}}$
4

Divide fractions $\frac{1}{\frac{1}{\sqrt{x^2+4}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$-\left(y-1\right)^2=\sqrt{x^2+4}$
5

Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$

$\int-\left(y^2-2y+1\right)dy=\int\sqrt{x^2+4}dx$

The trinomial $\left(y^2-2y+1\right)$ is a perfect square trinomial, because it's discriminant is equal to zero

$\Delta=b^2-4ac=-2^2-4\left(1\right)\left(1\right) = 0$

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{y^2}\:and\:b=\sqrt{1}$

Factoring the perfect square trinomial

$\int-\left(y-1\right)^{2}dy$

We can solve the integral $\int-\left(y-1\right)^{2}dy$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $y-1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=y-1$

Now, in order to rewrite $dy$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=dy$

Substituting $u$ and $dy$ in the integral and simplify

$\int-u^{2}du$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$-\int u^{2}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$\frac{-u^{3}}{3}$

Replace $u$ with the value that we assigned to it in the beginning: $y-1$

$\frac{-\left(y-1\right)^{3}}{3}$
6

Solve the integral $\int-\left(y^2-2y+1\right)dy$ and replace the result in the differential equation

$\frac{-\left(y-1\right)^{3}}{3}=\int\sqrt{x^2+4}dx$

We can solve the integral $\int\sqrt{x^2+4}dx$ by applying integration method of trigonometric substitution using the substitution

$x=2\tan\left(\theta \right)$

Now, in order to rewrite $d\theta$ in terms of $dx$, we need to find the derivative of $x$. We need to calculate $dx$, we can do that by deriving the equation above

$dx=2\sec\left(\theta \right)^2d\theta$

Substituting in the original integral, we get

$\int2\sqrt{4\tan\left(\theta \right)^2+4}\sec\left(\theta \right)^2d\theta$

Factor the polynomial $4\tan\left(\theta \right)^2+4$ by it's greatest common factor (GCF): $4$

$\int2\sqrt{4\left(\tan\left(\theta \right)^2+1\right)}\sec\left(\theta \right)^2d\theta$

The power of a product is equal to the product of it's factors raised to the same power

$\int4\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$

Applying the trigonometric identity: $1+\tan\left(\theta \right)^2 = \sec\left(\theta \right)^2$

$\int4\sqrt{\sec\left(\theta \right)^2}\sec\left(\theta \right)^2d\theta$
Why is tan(x)^2+1 = sec(x)^2 ?

The integral of a function times a constant ($4$) is equal to the constant times the integral of the function

$4\int\sqrt{\sec\left(\theta \right)^2}\sec\left(\theta \right)^2d\theta$

Simplify $\sqrt{\sec\left(\theta \right)^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$

$4\int\sec\left(\theta \right)^{2\frac{1}{2}}\sec\left(\theta \right)^2d\theta$

Multiply $2$ times $\frac{1}{2}$

$4\int\sec\left(\theta \right)\sec\left(\theta \right)^2d\theta$

When multiplying exponents with same base you can add the exponents: $\sec\left(\theta \right)\sec\left(\theta \right)^2$

$4\int\sec\left(\theta \right)^{3}d\theta$

Rewrite $\sec\left(\theta \right)^{3}$ as the product of two secants

$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$

When multiplying exponents with same base you can add the exponents: $\sec\left(\theta \right)^2\sec\left(\theta \right)$

$4\int\sec\left(\theta \right)^{3}d\theta$
7

Solve the integral $\int\sqrt{x^2+4}dx$ and replace the result in the differential equation

$\frac{-\left(y-1\right)^{3}}{3}=4\int\sec\left(\theta \right)^{3}d\theta$

Simplify the integral $\int\sec\left(\theta \right)^{3}d\theta$ applying the reduction formula, $\displaystyle\int\sec(x)^{n}dx=\frac{\sin(x)\sec(x)^{n-1}}{n-1}+\frac{n-2}{n-1}\int\sec(x)^{n-2}dx$

$4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{3-1}+\frac{3-2}{3-1}\int\sec\left(\theta \right)d\theta\right)$

Simplify the expression inside the integral

$4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)$

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)\right)$

Apply the trigonometric identity: $\sin\left(\theta \right)\sec\left(\theta \right)^n$$=\tan\left(\theta \right)\sec\left(\theta \right)^{\left(n-1\right)}$, where $x=\theta $ and $n=2$

$4\left(\frac{\tan\left(\theta \right)\sec\left(\theta \right)}{2}+\frac{1}{2}\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)\right)$

Express the variable $\theta$ in terms of the original variable $x$

$4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right)\right)$

The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors

$L.C.M.=2$

Combine and simplify all terms in the same fraction with common denominator $2$

$4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)$

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$
8

Solve the integral $4\int\sec\left(\theta \right)^{3}d\theta$ and replace the result in the differential equation

$\frac{-\left(y-1\right)^{3}}{3}=4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$

Final Answer

$\frac{-\left(y-1\right)^{3}}{3}=4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Linear Differential EquationExact Differential EquationSeparable Differential EquationHomogeneous Differential Equation

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Function Plot

Plotting: $\frac{1}{\left(y-1\right)^2}dx+\frac{1}{\sqrt{x^2+4}}dy$

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e
π
ln
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log
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Dx
|◻|
θ
=
>
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>=
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sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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