Find the particular solution of the given differential equation that satisfies the indicated condition: $\frac { d y } { d x } = 28 x ^ { 3 } y ^ { 2 }$ ; y = 6 when x = 1.
A) $y = \frac { 6 } { 43 - 42 x ^ { 4 } }$
B) $y = \frac { 6 } { 43 + 42 x ^ { 4 } }$
C) $y = - \frac { 6 } { 43 + 42 x ^ { 4 } }$
D) $y = \frac { 6 } { 43 - 42 x ^ { 3 } }$

Question

Quizplus · Accepted Answer

This differential equation is separable. Separating variables and integrating both sides gives $\int y^{-2} dy = \int 28x^3 dx$, leading to $-y^{-1} = 7x^4 + C$. Using the initial condition $y(1) = 6$, we find $C = -\frac{1}{6} + 7$. Thus, $y = \frac{6}{43 - 42x^4}$.

Find the Particular Solution of the Given Differential Equation That dydx=28x3y2\frac { d y } { d x } = 28 x ^ { 3 } y ^ { 2 }dxdy​=28x3y2

Find the Particular Solution of the Given Differential Equation That $\frac { d y } { d x } = 28 x ^ { 3 } y ^ { 2 }$