Question

#JEE Mains




(•,•)=> The degree order of the differential equation
$\bf{ \huge{ \red {\tt{y_{2} ^ {(3/2) }- y_{1} ^ {(1/2) }- 4 = 0}}}}$
is ?¿
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Answer 1

Answer:

Here is your answer for the Differential Equations Degree and Order

Step-by-step explanation:

$\infty \wp$

thank you

Answer 2

$\large\underline{\sf{Given- }}$

The differential equation is

$\rm :\longmapsto\: {\bigg(y_2 \bigg) }^{\dfrac{3}{2} } - {\bigg(y_1\bigg) }^{\dfrac{1}{2} } - 4 = 0$

$\large\underline{\sf{To\:Find - }}$

Degree and order of given Differential equation

Basic Concept Used :-

Order of Differential equation :- It is defined as higher order derivative exist in the given Differential equation.

Degree of Differential equation :- It is defined as the power of higher order derivative exist in the given Differential equation when all the other terms are free from radicals.

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\: {\bigg(y_2 \bigg) }^{\dfrac{3}{2} } - {\bigg(y_1\bigg) }^{\dfrac{1}{2} } - 4 = 0$

Since, higher order derivative exist in the given Differential equation is 2.

$\bf\implies \:Order \: of \: differential \: equation \: is \: 2$

Now, to find the degree of Differential equation, we have to remove the radicals first from the given equation.

So, given equation can be rewritten as

$\rm :\longmapsto\: {\bigg(y_2 \bigg) }^{\dfrac{3}{2} } = {\bigg(y_1\bigg) }^{\dfrac{1}{2} } + 4$

On squaring both sides, we get

$\rm :\longmapsto\: {y_2}^{3} = {y_1}^{1} + 16 - 8{\bigg(y_1 \bigg) }^{\dfrac{1}{2} }$

$\red{\bigg \{ \because \tt \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy\bigg \}}$

and

$\red{\bigg \{ \because \tt \: {( {x}^{m}) }^{n} \: = \: {x}^{mn} \bigg \}}$

$\rm :\longmapsto\: {y_2}^{3} = {y_1}^{} + 16 - 8{\bigg(y_1 \bigg) }^{\dfrac{1}{2} }$

can be rewritten as

$\rm :\longmapsto\: {y_2}^{3} - {y_1}^{} - 16 = - 8{\bigg(y_1 \bigg) }^{\dfrac{1}{2} }$

We know,

$\boxed{ \rm{ {(a + b + c)}^{2}={a}^{2} + {b}^{2} +{c}^{2} + 2ab + 2bc + 2ca}}$

Now, Squaring both sides, we get

$\rm :\longmapsto\: {y_2}^{6} + {y_1}^{2} + 256 - 2y_1 {y_2}^{3} + 32y_1 - 32 {y_2}^{3} = 64y_1$

$\rm :\longmapsto\: {y_2}^{6} + {y_1}^{2} + 256 - 2y_1 {y_2}^{3} + 32y_1 - 32 {y_2}^{3} - 64y_1 = 0$

$\rm :\longmapsto\: {y_2}^{6} + {y_1}^{2} + 256 - 2y_1 {y_2}^{3} - 32y_1 - 32 {y_2}^{3} = 0$

can be written in rearranged manner as

$\rm :\longmapsto\: {y_2}^{6} - {32y_2}^{3} - 2y_1 {y_2}^{3} + {y_1}^{2} - 32y_1 + 256 = 0$

Now, we found that all the terms are free from radicals.

So, degree of higher order derivative exist in the given Differential equation is 6.

$\bf\implies \:Degree \: of \: differential \: equation \: is \: 6$