Show that the differential equation  is homogeneous. Find the

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 Multiple Choice QuestionsShort Answer Type

311. If space straight y space equals space Pe to the power of ax plus Qe to the power of bx show that
straight d to the power of 2 straight y end exponent over dx squared minus left parenthesis straight a plus straight b right parenthesis dy over dx plus aby space equals 0
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312. If space straight x equals cost left parenthesis 3 minus 2 cos squared straight t right parenthesis space and space straight y equals sint left parenthesis 3 minus 2 sin squared straight t right parenthesis comma space find space the space value space of space dy over dx at space straight t equals straight pi over 4.
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313.

Find the particular solution of the differential equation log open parentheses dy over dx close parentheses equals 3 straight x plus 4 straight y comma space given space that space straight y space equals space 0 comma space when space straight x equals space 0.

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314.

Write the degree of the differential equation straight x cubed open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses squared space plus space open parentheses dy over dx close parentheses to the power of 4 space equals space 0

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315.

The amount of pollution content added in air in city due to x-diesel vehicles is given by straight P left parenthesis straight x right parenthesis space equals space 0.005 straight x cubed plus 0.02 straight x squared plus 30 straight x. Find the marginal increase in pollution content when 3 diesel vehicles are added and write which value is indicated in the above questions.

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316.

Show that the function f in straight A space equals space straight R to the power of minus open curly brackets 2 over 3 close curly brackets space defined space as space straight f left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction space is space one minus one space and space onto. space Hence space find space straight f to the power of negative 1 end exponent.

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317.

Differentiate the following function with respect to x:
left parenthesis log space straight x right parenthesis to the power of straight x plus straight x to the power of log straight x

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318. If space straight y space equals space log space open square brackets straight x space plus space square root of straight x squared plus space straight a squared end root close square brackets comma space show space that space left parenthesis straight x squared space plus space straight a squared right parenthesis space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus straight x dy over dx space equals space 0
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319.

If straight x equals straight a space sint space and space straight y space equals space straight a open parentheses cost plus log space tan straight t over 2 close parentheses space find space fraction numerator straight d squared straight y over denominator dx squared end fraction

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 Multiple Choice QuestionsLong Answer Type

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320.

Show that the differential equation 2 ye to the power of straight x divided by straight y end exponent dx plus left parenthesis straight y minus 2 straight x space straight e to the power of straight x divided by straight y end exponent right parenthesis space dy space equals space 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.


2 ye to the power of straight x divided by straight y end exponent dx plus left parenthesis straight y minus 2 straight x space straight e to the power of straight x divided by straight y end exponent right parenthesis dy space equals space 0
dy over dx space equals space fraction numerator 2 xe to the power of begin display style straight x over straight y end style end exponent minus straight y over denominator 2 ye to the power of begin display style straight x over straight y end style end exponent end fraction space space space space... left parenthesis 1 right parenthesis
Let space straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator 2 xe to the power of begin display style straight x over straight y end style end exponent minus straight y over denominator 2 ye to the power of begin display style straight x over straight y end style end exponent end fraction
 Then comma space straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda open parentheses 2 xe to the power of begin display style straight x over straight y end style end exponent minus straight y close parentheses over denominator straight lambda open parentheses 2 ye to the power of begin display style straight x over straight y end style end exponent close parentheses end fraction equals straight lambda degree open square brackets straight F left parenthesis straight x comma space straight y right parenthesis close square brackets
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. 
Let x = vy
Differentiating w.r.t. y, we get
dx over dy equals straight v plus straight y dv over dy
Substituting the value of x and dx over dy in equation (1), we get
straight v plus straight y dv over dy equals fraction numerator 2 vye to the power of straight v minus straight y over denominator 2 ye to the power of straight v end fraction space equals fraction numerator 2 ve to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction
or space space straight y dv over dy space equals fraction numerator 2 ve to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction minus straight v
or space space straight y dv over dy equals negative fraction numerator 1 over denominator 2 straight e to the power of straight v end fraction
or space space 2 straight e to the power of straight v dv space equals fraction numerator negative dy over denominator straight y end fraction
or space space integral 2 straight e to the power of straight v. dv space equals space minus integral dy over straight y
or space space 2 straight e to the power of straight v space equals space minus log space open vertical bar straight y close vertical bar plus straight C
Substituting the value of v, we get
2 straight e to the power of straight x over straight y end exponent plus log space open vertical bar straight y close vertical bar space equals space straight C space space... left parenthesis 2 right parenthesis
Substituting x = 0 and y = 1 in equation (2), we get
2 straight e degree plus log space open vertical bar 1 close vertical bar space equals space straight C space rightwards double arrow space straight C space equals space 2
Substituting the value of C in equation (2), we get
2 straight e to the power of straight x over straight y end exponent plus log space open vertical bar straight y close vertical bar space equals space 2 comma which is the particular solution of the given differential equation. 

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