Find the particular solution for the differential equation \(\frac{dy}{dx}=\frac{3x^2}{7y}\) given that, y=1 when x=1.

Category: QuestionsFind the particular solution for the differential equation \(\frac{dy}{dx}=\frac{3x^2}{7y}\) given that, y=1 when x=1.
Editor">Editor Staff asked 11 months ago

Find the particular solution for the differential equation \(\frac{dy}{dx}=\frac{3x^2}{7y}\) given that, y=1 when x=1.
 
(a) 7x^2=2y^3+5
 
(b) 7x^3=2y^2+5
 
(c) 7y^2=2x^3+5
 
(d) 2y^2=5x^3+6
 
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This interesting question is from Methods of Solving First Order & First Degree Differential Equations topic in portion Differential Equations of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

The correct answer is (c) 7y^2=2x^3+5
 
To explain I would say: Given that, \(\frac{dy}{dx}=\frac{3x^2}{7y}\)
 
Separating the variables, we get
 
7y dy=3x^2 dx
 
Integrating both sides, we get
 
\(7\int y \,dy=3\int x^2 \,dx\)
 
\(\frac{7y^2}{2}=3(\frac{x^3}{3})+C\)
 
\(\frac{7y^2}{2}=x^3\)+C –(1)
 
Given that y=1, when x=1
 
Substituting the values in equation (1), we get
 
\(\frac{7(1)^2}{2}=(1)^3+C\)
 
\(C=\frac{7}{2}-1=\frac{5}{2}\)
 
Hence, the particular solution of the given differential equation is:
 
\(\frac{7y^2}{2}=x^3+\frac{5}{2}\)
 
⇒7y^2=2x^3+5