Question

D.E. whose solution is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] :
\[\left( 1 \right)\] \[xy{y_2} + xy_1^2 = y{y_1}\]
\[\left( 2 \right)\] \[xy{y_2} + {y_1} = y\]
\[\left( 3 \right)\] \[{x^2}{y_2} + x{y_1} = y\]
\[\left( 4 \right)\] \[{x^2}{y_2} + 2x{y_1} = y_1^2\]

Answer 1

Hint: We have to find the differential equation for the given solution . We solve this question using the concept of the differentiation of variables . We will first differentiate the given solution with respect to \[x\] and then simplify the equation such that we get the value of the constant terms in terms of an expression of derivatives . Then again differentiating the expression with respect to \[x\] and simplifying the expression , we will get the required differential equation for the given solution .

Complete step-by-step answer:
Given :
Solution for the differential equation is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] .
Now we have to differentiate the given expression with respect to \[x\] .
Taking L.C.M. and simplifying the expression , we can write it as :
\[{b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2} - - - \left( 1 \right)\]
As we know that the formulas for derivatives are given as :
Derivative of \[{x^n} = n \times {x^{^{n - 1}}}\]
Derivative of \[constant = 0\]
Using the formulas of derivatives and differentiating equation \[\left( 1 \right)\] with respect to \[x\] , we can write the expression as :
\[2{b^2}x + 2{a^2}y{y_1} = 0\]
Where \[{y_1}\] is the first derivative of the function \[y\] with respect to \[x\] . i.e. \[{y_1} = \dfrac{{dy}}{{dx}}\]
Now , simplifying the expression , we can write the expression as :
\[2{b^2}x = - 2{a^2}y{y_1}\]
\[\dfrac{{{b^2}}}{{{a^2}}} = \dfrac{{ - y{y_1}}}{x} - - - \left( 2 \right)\]
Now , we will differentiate the equation \[\left( 2 \right)\] with respect to \[x\] .
Also we know that the quotient rule for derivatives is given as :
\[\dfrac{d}{{dx}}\left( {\dfrac{f}{g}} \right) = \dfrac{{\dfrac{d}{{dx}}f \times g - f \times \dfrac{d}{{dx}}g}}{{{g^2}}}\]
Also we know that the quotient rule for derivatives is given as :
\[\dfrac{d}{{dx}}\left( {f \times g} \right) = \dfrac{d}{{dx}}f \times g + f \times \dfrac{d}{{dx}}g\]
Using the formulas of derivatives and differentiating equation \[\left( 2 \right)\] with respect to \[x\] , we can write the expression as :
\[\dfrac{{\dfrac{d}{{dx}}\left( { - y{y_1}} \right) \times x - \left( { - y{y_1}} \right)\dfrac{d}{{dx}}x}}{{{x^2}}} = 0\]
Also , we can write the expression as :
\[\dfrac{{\left( { - y{y_2} - y_1^2} \right) \times x + y{y_1}}}{{{x^2}}} = 0\]
On further simplifying , we can write the expression as :
\[ - xy{y_2} - xy_1^2 + y{y_1} = 0\]
\[xy{y_2} + xy_1^2 = y{y_1}\]
Where \[{y_2}\] is the second derivative of the function \[y\] with respect to \[x\] . i.e. \[{y_2} = \dfrac{{{d^2}y}}{{d{x^2}}}\] .
Hence , the differential equation for the given solution \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] is \[xy{y_2} + xy_1^2 = y{y_1}\] .
So, the correct answer is “Option 1”.

Note: For finding the differential equation of a given solution , we have to eliminate the constant terms of the expression of the solution . We differentiate the solution according to the number of constants in the solution . In this question we had two constants so we differentiated the solution two times .