Find out the answer to "Classify the following differential equations as separable, homogeneous,..."

Inyalan0

Answered question

2021-12-30

Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification.

2 x \sin^{2} y d x - (x^{2} - 9) \cos y d y = 0

Answer & Explanation

Philip Williams

Beginner2021-12-31Added 39 answers

Given differential equation

2 x \sin^{2} y d x - (x^{2} - 9) \cos y d y = 0

write the equation in the standard form

\frac{d y}{d x} = \frac{2 x \sin^{2} y}{(x^{2} - 9) \cos y}

we can saperate the variable so, equation is seperable now seperating the variable

(\frac{\cos y}{\sin^{2} y}) d y = (\frac{2 x}{x^{2} - 9}) d x

integrate both side

\Rightarrow \int (\frac{\cos y}{\sin^{2} y}) d y = \int (\frac{2 x}{x^{2} - 9}) d x

\Rightarrow \int \frac{d (\sin y)}{\sin^{2} y} = \int \frac{d (x^{2} - 9)}{x^{2} - 9}

\Rightarrow - \frac{1}{\sin y} = \ln (x^{2} - 9) + C

\Rightarrow \ln (x^{2} - 9) + \cos e c y = C

hence solution of differential equation is

\ln (x^{2} - 9) + \cos e c y = C

temnimam2

Beginner2022-01-01Added 36 answers

2 x \sin^{2} y d x - (x^{2} - 9) \cos y d y = 0

∴ 2 x \sin^{2} y d x = (x^{2} - 9) \cos y d y

∴ \frac{2 x}{(x^{2} - 9)} d x = \frac{\cos y}{\sin^{2} y} d y

∴ \frac{2 x}{(x^{2} - 9)} d x = \cos e c y \cdot \cot y d y

This is variable separable form.

∴

the general solution is

\int \frac{2 x}{x^{2} - 9} d x = \int \cos e c y \cdot \cot y d y

∴ \ln | x^{2} - 9 | = - \cos e c y + c

This is general solution.

karton

Expert2022-01-09Added 613 answers

$\begin{matrix} \frac{d y}{d x} = \frac{2 x \sin^{2} y}{(x^{2} - 9) \cos y} \\ (\frac{\cos y}{\sin^{2} y}) d y = (\frac{2 x}{x^{2} - 9}) d x \\ \int (\frac{\cos y}{\sin^{2} y}) d y = \int (\frac{2 x}{x^{2} - 9}) d x \\ \int \frac{d (\sin y)}{\sin^{2} y} = \int \frac{d (x^{2} - 9)}{x^{2} - 9} \\ - \frac{1}{\sin y} = \ln (x^{2} - 9) + C \\ \ln (x^{2} - 9) + \cos e c y = C \\ A n s w e r : \ln (x^{2} - 9) + cosec y = C \end{matrix}$

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