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Differential Equations

Short Answer Type

141.

Solve:
$open parentheses straight e to the power of straight x plus 1 close parentheses space straight y space dy space plus space left parenthesis straight y plus 1 right parenthesis space straight e to the power of straight x space dx space equals space 0$

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142.
Solve:
$straight y left parenthesis 1 minus straight x squared right parenthesis space dy space plus space straight x space left parenthesis 1 plus straight y squared right parenthesis space dx space equals space 0.$

The given differential equation is

straight y left parenthesis 1 minus straight x squared right parenthesis space dy space plus space straight x space left parenthesis 1 plus straight y squared right parenthesis space dx space equals space 0

straight y left parenthesis 1 minus straight x squared right parenthesis space dy space equals space minus straight x left parenthesis 1 plus straight y squared right parenthesis space dx

fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus fraction numerator straight x over denominator 1 minus straight x squared end fraction dx

Integrating,

integral fraction numerator straight y over denominator 1 plus straight y squared end fraction dy equals space space minus integral fraction numerator straight x over denominator 1 minus straight x squared end fraction dx

therefore space space space space integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space integral space fraction numerator negative 2 straight x over denominator 1 minus straight x squared end fraction dx therefore space space log space left parenthesis 1 plus straight y squared right parenthesis space equals space log space open vertical bar 1 minus straight x squared close vertical bar plus straight c

which is required solution.

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143.

Solve:
$straight y space logydx space minus space straight x space dy space equals space 0$

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Long Answer Type

144.

Show that the general solution of the differential equation $dy over dx plus fraction numerator straight y squared plus straight y plus 1 over denominator straight x squared plus straight x plus 1 end fraction space equals 0$ is given by (x + y + 1) = A (1 – x – y – 2 x y), where A is parameter.

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Short Answer Type

145.

Solve:
$dy over dx space equals space fraction numerator straight x left parenthesis 2 space log space straight x space plus space 1 right parenthesis over denominator sin space straight y space plus space straight y space cosy end fraction$

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146.

Solve
$dy over dx space equals space fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x space cosy end fraction$

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Long Answer Type

147.

Solve
$dy over dx space equals space sin cubed straight x space cos squared straight x plus straight x space straight e to the power of straight x$

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148.

Solve:
$dy over dx space equals space cos cubed straight x space sin to the power of 4 straight x plus straight x square root of 2 straight x plus 1 end root$

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149.

Solve:
$dy over dx equals negative straight x space sin squared straight x space space equals space fraction numerator 1 over denominator straight x space log space straight x end fraction$