Question

solution of x^2p+y^2q=z^2​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The solution of

$\sf{ {x}^{2}p + {y}^{2} q = {z}^{2} \: \: }$

EVALUATION

The given equation is

$\sf{ {x}^{2}p + {y}^{2} q = {z}^{2} \: \: }$

Here the subsidiary equations are

$\displaystyle \sf{ \frac{dx}{ {x}^{2} } = \frac{dy}{ {y}^{2} } = \frac{dz}{ {z}^{2} }\: } \: \: \: \: \: .....(1)$

From first two of (1)

$\displaystyle \sf{ \frac{dx}{ {x}^{2} } = \frac{dy}{ {y}^{2} } }$

On integration

$\displaystyle \sf{ - \frac{1}{x} = - \frac{1}{y} + a }$

$\implies \displaystyle \sf{ \frac{1}{y} - \frac{1}{x} = a } \: \: \: ....(2)$

Again from last two of (1)

$\displaystyle \sf{ \frac{dy}{ {y}^{2} } = \frac{dz}{ {z}^{2} }\: } \: \: \: \: \:$

On integration

$\displaystyle \sf{ - \frac{1}{y} = - \frac{1}{z} + b}$

$\implies\displaystyle \sf{ \frac{1}{z} - \frac{1}{y} = b} \: \: \: .....(3)$

Where a and b are constants

From Equation (2) & Equation (3) we get the required solution as

$\displaystyle \sf{ \frac{1}{y} - \frac{1}{x} = f \bigg( \frac{1}{z} - \frac{1}{x} \bigg)\: }$

RESULT

The required solution is

$\boxed{ \displaystyle \sf{ \: \: \: \: \frac{1}{y} - \frac{1}{x} = f \bigg( \frac{1}{z} - \frac{1}{x} \bigg)\: } \: \: \: \: }$

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Answer 2

Step-by-step explanation:

\displaystyle\huge\red{\underline{\underline{Solution}}}

Solution

TO DETERMINE

The solution of

\sf{ {x}^{2}p + {y}^{2} q = {z}^{2} \: \: }x

2

p+y

2

q=z

2

EVALUATION

The given equation is

\sf{ {x}^{2}p + {y}^{2} q = {z}^{2} \: \: }x

2

p+y

2

q=z

2

Here the subsidiary equations are

\displaystyle \sf{ \frac{dx}{ {x}^{2} } = \frac{dy}{ {y}^{2} } = \frac{dz}{ {z}^{2} }\: } \: \: \: \: \: .....(1)

x

2

dx

=

y

2

dy

=

z

2

dz

.....(1)

From first two of (1)

\displaystyle \sf{ \frac{dx}{ {x}^{2} } = \frac{dy}{ {y}^{2} } }

x

2

dx

=

y

2

dy

On integration

\displaystyle \sf{ - \frac{1}{x} = - \frac{1}{y} + a }−

x

1

=−

y

1

+a

\implies \displaystyle \sf{ \frac{1}{y} - \frac{1}{x} = a } \: \: \: ....(2)⟹

y

1

−

x

1

=a....(2)

Again from last two of (1)

\displaystyle \sf{ \frac{dy}{ {y}^{2} } = \frac{dz}{ {z}^{2} }\: } \: \: \: \: \:

y

2

dy

=

z

2

dz

On integration

\displaystyle \sf{ - \frac{1}{y} = - \frac{1}{z} + b}−

y

1

=−

z

1

+b

\implies\displaystyle \sf{ \frac{1}{z} - \frac{1}{y} = b} \: \: \: .....(3)⟹

z

1

−

y

1

=b.....(3)

Where a and b are constants

From Equation (2) & Equation (3) we get the required solution as

\displaystyle \sf{ \frac{1}{y} - \frac{1}{x} = f \bigg( \frac{1}{z} - \frac{1}{x} \bigg)\: }

y

1

−

x

1

=f(

z

1

−

x

1

)

RESULT

The required solution is

\boxed{ \displaystyle \sf{ \: \: \: \: \frac{1}{y} - \frac{1}{x} = f \bigg( \frac{1}{z} - \frac{1}{x} \bigg)\: } \: \: \: \: }

y

1

−

x

1

=f(

z

1

−

x

1

)

━━━━━━━━━━━━━━━━

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Solve the differential equation

(cosx.cosy - cotx) dx - (sinx.siny) dy=0

https://brainly.in/question/23945760