Question

1. The
solution of the difficult
equation
cos^2 x d^2y/dx^2 = 1 is​

Answer 1

Step-by-step explanation:

We have,

$\cos^{2} (x) . \frac{ {d}^{2} y}{d {x}^{2} } = 1$

$\implies \frac{ {d}^{2} y}{d {x}^{2} } = \sec^{2} (x)$

$\implies \: d( \frac{dy}{dx} ) = \sec^{2} (x) dx$

Integrating both sides,

$\implies \: \int \: d( \frac{dy}{dx} ) = \int\sec^{2} (x) dx$

$\implies \frac{dy}{dx} = \tan(x) + c_{1}$

$\implies \: dy = ( \tan(x) + c _{1})dx$

Integrating both sides,

$\implies \: \int dy = \int( \tan(x) + c _{1})dx$

$\implies \: y = ln( \sec(x) ) + c_{1}x + c_{2}$

Answer 2

Answer:

We have,

\begin{gathered} \cos^{2} (x) . \frac{ {d}^{2} y}{d {x}^{2} } = 1 \\ \end{gathered}

cos

2

(x).

dx

2

d

2

y

=1

\begin{gathered} \implies \frac{ {d}^{2} y}{d {x}^{2} } = \sec^{2} (x) \\ \end{gathered}

⟹

dx

2

d

2

y

=sec

2

(x)

\begin{gathered} \implies \: d( \frac{dy}{dx} ) = \sec^{2} (x) dx \\ \end{gathered}

⟹d(

dx

dy

)=sec

2

(x)dx

Integrating both sides,

\begin{gathered} \implies \: \int \: d( \frac{dy}{dx} ) = \int\sec^{2} (x) dx \\ \end{gathered}

⟹∫d(

dx

dy

)=∫sec

2

(x)dx

\begin{gathered} \implies \frac{dy}{dx} = \tan(x) + c_{1} \\ \end{gathered}

⟹

dx

dy

=tan(x)+c

1

\begin{gathered}\implies \: dy = ( \tan(x) + c _{1})dx \\ \end{gathered}

⟹dy=(tan(x)+c

1

)dx

Integrating both sides,

\begin{gathered}\implies \: \int dy = \int( \tan(x) + c _{1})dx \\ \end{gathered}

⟹∫dy=∫(tan(x)+c

1

)dx

\begin{gathered}\implies \: y = ln( \sec(x) ) + c_{1}x + c_{2} \\ \end{gathered}

⟹y=ln(sec(x))+c

1

x+c

2