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Solve the following differential equation: x cos y dy = ex(x log x + 1) dx

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प्रश्न

Solve the following differential equation:

x cos y dy = ex(x log x + 1) dx

योग
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उत्तर

The equation can be written as

cos y dy = `"e"^x ((xlogx + 1))/x  "d"x`

cos y dy = `"e"^x [(xlogx)/x + 1/x]  "d"x`

cos y dy = `"e"^x [logx + 1/x]  "d"x`

Taking integration on both sides, we get

`int cos y  "d"y = int "e"^x [log x + 1/x]  "d"x`  ........(1)

R.H.S

`int "e"^x [log x + 1/x]  "d"x`

⇒ Take f(x) = log x

f'(x) = `1/x`

This of the form `int "e"^x  ["f"(x) + "f'"(x)]  "d"x = "e"^x  "f"(x) + "C"`

∴ `int "e"^x [log x + 1/x]  "d"x = "e"^x lo x + "C"`

Substituting in (1), we get

sin y = ey log x + C

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Solution of First Order and First Degree Differential Equations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Ordinary Differential Equations - Exercise 10.5 [पृष्ठ १६२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 10 Ordinary Differential Equations
Exercise 10.5 | Q 4. (viii) | पृष्ठ १६२

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