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Differential Equations

Short Answer Type

141.

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

87 Views

142.

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

83 Views

143.

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

91 Views

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.

77 Views

Short Answer Type

145.

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

80 Views

146.
Solve
$dy over dx space equals space fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x space cosy end fraction$

The given differential equation is $dy over dx equals fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x space cosy end fraction$
Separating the variables, we get,
$cosy space dy space equals space fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x end fraction dx space space or space space cosy space dy space equals space straight e to the power of straight x open parentheses fraction numerator straight x space logx space plus space 1 over denominator straight x end fraction close parentheses dx$
or $cosy space dy space equals space straight e to the power of straight x open parentheses logx plus 1 over straight x close parentheses dx space space space rightwards double arrow space space integral space cosy space dy space equals space integral straight e to the power of straight x open parentheses logx plus 1 over straight x close parentheses dx$
$therefore space space space sin space straight y space equals space straight e to the power of straight x space logx space plus space straight c space space space space space space space space space open square brackets because space space integral straight e to the power of straight x open curly brackets straight f left parenthesis straight x right parenthesis plus straight f apostrophe left parenthesis straight x right parenthesis close curly brackets dx space equals space straight e to the power of straight x space straight f left parenthesis straight x right parenthesis space plus space straight c close square brackets$
which is required solution.

83 Views

Long Answer Type

147.

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

82 Views

148.

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

84 Views

149.

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