Test Index

CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

© examsnet.com
Question : 20 of 36
Marks: +1, -0
Show that the function y=ax+2a2y = a x + 2 a^2 is a solution of the differential equation 2(dydx)2+x(dydx)−y=02 \left(\frac{dy}{dx}\right)^2 + x \left(\frac{dy}{dx}\right) - y = 0
Solution:     👈: Video Solution
The given function is
y=ax+2a2⋅⋅⋅⋅⋅⋅⋅(i)y = a x + 2 a^2 \quad \cdot\cdot\cdot\cdot\cdot\cdot\cdot \text{(i)}
Differentiating equation (i) w.r.t. ' xx ', we get dydx=a\frac{dy}{dx} = a
Substituting the value of 'a' in equation, we get
y=x(dydx)+2(dydx)2y = x \left(\frac{dy}{dx}\right) + 2 \left(\frac{dy}{dx}\right)^2
⇒2(dydx)2+x(dydx)−y=0\Rightarrow 2 \left(\frac{dy}{dx}\right)^2 + x \left(\frac{dy}{dx}\right) - y = 0
© examsnet.com
Go to Question:
Prev Question
Next Question