# 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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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