Which of the following differential equations has the family of solutions&nbsp;x2+cy2=?a)x2y′+y+x=0b)&nbsp;xy+(1−x2)y′=0&nbsp;c)&nbsp;x2y′−xy=1d)&nbsp;x2y+(1-x2)y'=0

Step 1Given,family of solutionsx2+cy2=1 ...(1)We went to find its differential equationStep 2Now,diff (1) w.r.t x we get2x+c×2yy′=0⇒x+cyy′=0⇒c=−xyy′ ...(1)putting the value of c from eqn (2) into eqn (1)(1)⇒x2+(−xyy′)y2=1⇒x2−xyy′=1⇒x2y′−xy=y′⇒y′−x2y′+xy=0⇒y′(1−x2)+xy=0r.c. xy+(1−x2)+y′=0This xy+(1−x2)y′=0 is the requred diff eqnfor given family of solutions (1)Thefore obtion (B) is correct

Explained: "Which of the following differential equations has the..."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

Lawson Herrera

Answered question

2022-03-13

Which of the following differential equations has the family of solutions $x^{2} + c y^{2} = ?$
a) $x^{2} y^{'} + y + x = 0$
b) $x y + (1 - x^{2}) y^{'} = 0$
c) $x^{2} y^{'} - x y = 1$
d) $x^{2} y + (1 - x^{2}) y' = 0$

Answer & Explanation

Kyson Sloan

Beginner2022-03-14Added 5 answers

Step 1
Given,family of solutions

x^{2} + c y^{2} = 1

...(1)
We went to find its differential equation
Step 2
Now,diff (1) w.r.t x we get

2 x + c \times 2 y y^{'} = 0

\Rightarrow x + c y y^{'} = 0

\Rightarrow c = \frac{- x}{y y^{'}}

...(1)
putting the value of c from

e q^{n}

(2) into

e q^{n}

(1)
(1)

\Rightarrow x^{2} + (\frac{- x}{y y^{'}}) y^{2} = 1

\Rightarrow x^{2} - \frac{x}{y y^{'}} = 1

\Rightarrow x^{2} y^{'} - x y = y^{'}

\Rightarrow y^{'} - x^{2} y^{'} + x y = 0

\Rightarrow y^{'} (1 - x^{2}) + x y = 0

r.c.

x y + (1 - x^{2}) + y^{'} = 0

This

x y + (1 - x^{2}) y^{'} = 0

is the requred diff

e q^{n}

for given family of solutions (1)
Thefore obtion (B) is correct

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Pre-Algebra

Didn't find what you were looking for?