Question

Find a general expression for y given that
a
dy
-
=
9x – 2
3x
dx


I need help with either question 2 or 3 please!

Answer 1

Answer:

See Below.

Step-by-step explanation:

Problem #2

We want to find the general expression for the differential equation:

$\displaystyle \frac{dy}{dx} = \frac{9x-2}{3\sqrt{x}}$

Separation of variables:
$\displaystyle dy = \frac{9x-2}{3\sqrt{x}} \, dx$

Take the indefinite integral of both sides and integrate:
$\displaystyle \begin{aligned} \int 1 \, dy & = \int \frac{9x-2}{3\sqrt{x}} \, dx \\ \\ y & = \int \frac{9x}{3\sqrt{x}} - \frac{2}{3\sqrt{x}}\,dx \\ \\ & = \int 3 x\cdot x^{-1/2} \, dx +\int -\frac{2}{3}x^{-1/2}\, dx \\ \\ & = 3\int x^{1/2} \, dx -\frac{2}{3}\int x^{-1/2} \, dx \\ \\ & = 3\left(\frac{2}{3}x^{3/2}\right)-\frac{2}{3}\left(2x^{1/2}\right) \\ \\ & = 2x^{3/2} -\frac{4}{3}\sqrt{x} + C\end{aligned}$

Hence, the general expression is:
$\displaystyle y = 2x^{3/2} -\frac{4}{3}\sqrt{x} + C$

Problem #3

We want to evaluate the integral:

$\displaystyle \int \left( 5-x^2 + \frac{18}{x^4}\right) \, dx$

Apply the power rule for integrals:
$\displaystyle \begin{aligned} \int \left(5-x^2+\frac{18}{x^4}\right)\, dx & = \int \left(5x^0-x^2+18x^{-4}\right)\, dx \\ \\ & = 5x-\frac{1}{3}x^3+18\left(-\frac{1}{3}x^{-3}\right) + C \\ \\ & = 5x-\frac{1}{3}x^3 -\frac{18}{3}x^{-3} + C \\ \\ & = 5x-\frac{1}{3}x^3-\frac{6}{x^3} + C\end{aligned}$

In conclusion:
$\displaystyle \begin{aligned} \int \left(5-x^2+\frac{18}{x^4}\right)\, dx & = 5x-\frac{1}{3}x^3-\frac{6}{x^3} + C\end{aligned}$

Answer 2

We need to solve two questions first from differential equations and the second one from Integral Calculus, but before starting let's recall the formulae and concepts which will help us a lot in doing the questions:

• ${\boxed{\displaystyle \bf \int x^{n}\:dx=\dfrac{x^{n+1}}{n+1}+C\:\: , \:\: n\neq -1}}$

• ${\boxed{\displaystyle \bf \int \{f(x)\pm g(x)\pm h(x)\pm \cdots\}dx=\int f(x)\: dx\pm \int g(x)\:dx\pm \int h(x)\:dx\pm \cdots}}$

• ${\boxed{\displaystyle \bf \int kf(x)\:dx=k\int f(x)\:dx}}$

• ${\boxed{\displaystyle \bf \int dx=x+C}}$

Where , C is the Arbitrary Constant and k being any constant

Question 1 :-

Consider the differential equation :-

${:\implies \quad \sf \dfrac{dy}{dx}=\dfrac{9x-2}{3\sqrt{x}}}$

${:\implies \quad \sf dy=\dfrac{9x-2}{3\sqrt{x}}dx}$

Seperate the denominators of the fraction in RHS

${:\implies \quad \sf dy=\bigg(\dfrac{9x}{3\sqrt{x}}-\dfrac{2}{3\sqrt{x}}\bigg)dx}$

${:\implies \quad \sf dy=\bigg(3\sqrt{x}-\dfrac{2(x)^{\footnotesize -\dfrac12}}{3}\bigg)dx}$

Integrating both sides ;

${:\implies \quad \displaystyle \sf \int dy=\int \bigg(3\sqrt{x}-\dfrac{2(x)^{\footnotesize -\dfrac12}}{3}\bigg)dx}$

${:\implies \quad \displaystyle \sf y=\int 3(x)^{\footnotesize \dfrac12}\:dx-\int \dfrac{2(x)^{\footnotesize -\dfrac12}}{3}\:dx}$

${:\implies \quad \displaystyle \sf y=3\int (x)^{\footnotesize \dfrac12}\:dx-\dfrac23 \int (x)^{\footnotesize -\dfrac12}\:dx}$

${:\implies \quad \displaystyle \sf y=3\times\dfrac{(x)^{\footnotesize \dfrac32}}{\dfrac32}-\dfrac{2}{3}\times \dfrac{(x)^{\footnotesize \dfrac12}}{\dfrac12}+C}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=2x\sqrt{x}-\dfrac{4}{3}\sqrt{x}+C}}}$

Question 2 :-

Refer to the attachment for this answer