Given:
()2−x()+y=0 ......... (1)
These types of questions will be solved by option.
Checking Option 1: y=x−1 Differentiating with respect to x, we get
⇒=1−0 ⇒=1 Put the value of
=1 in equation 1st,
12−(x×1)+y=0 ⇒1−x+y=0 ⇒1−x+x−1=0 (∵y=x−1) ⇒0=0 Hence y = x - 1 is the solution for the differential equation.
Checking Option 2: 4y=x2Differentiating with respect to
x, we get
⇒4=2x ⇒= Put the value of
= in equation
1‌st ()2−x()+y=0 ⇒−+y=0 ⇒−x2+4y=0 ⇒−x2+x2=0 (∵4y=x2) ⇒0=0 Hence
4y=x2 is the solution for the differential equation.
Checking Option 3: y = x
Differentiating with respect to x, we get
Put the value of
=1 in equation
1st,
12−(x×1)+y=0 ⇒1−x+y=0 ⇒1−x+x=0‌‌‌(∵y=x) ⇒1≠0 Hence y = x is the not a solution for the differential equation.
Checking Option 4: y = -x - 1
Differentiating with respect to x, we get
⇒=−1−0 ⇒=−1 Put the value of
=−1 in equation 1 st,
−12−(x×−1)+y=0 ⇒1+x+y=0 ⇒1+x−x−1=0‌‌‌‌(∵y=−x−1) ⇒0=0 Hence y = -x - 1 is the solution for the differential equation.
So, option 3 is correct answer.