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प्रश्न
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उत्तर
We have,
\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{\cos x \sin y}{\cos y}\]
\[ \Rightarrow \frac{\cos y}{\sin y}dy = - \cos x dx\]
\[ \Rightarrow \cot y dy = - \cos x dx\]
Integrating both sides, we get
\[\int\cot y dy = - \int\cos x dx\]
\[ \Rightarrow \log \left| \sin y \right| = - \sin x + C\]
\[\text{ Hence,} \log \left| \sin y \right| = - \sin x +\text{ C is the required solution .} \]
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संबंधित प्रश्न
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} = y\]
|
y = ax |
xy dy = (y − 1) (x + 1) dx
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
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| Solution | D.E. |
| ax2 + by2 = 5 | `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx` |
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Choose the correct alternative:
Solution of the equation `x("d"y)/("d"x)` = y log y is
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
