For the differential equation, find the particular solution if $y = \frac { 1 } { 2 }$ when $x = \frac { 1 } { 5 }$ . $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } } { x ^ { 2 } }$
A) $y = \frac { 1 } { 1 - 3 x }$
B) $y = \frac { x } { 1 - 3 x }$
C) $y = \frac { x } { 3 - x }$
D) $y = - \frac { 3 } { 1 - x }$
E) $y = \frac { x } { 1 + 3 x }$

Question

For the differential equation, find the particular solution if $y = \frac { 1 } { 2 }$ when $x = \frac { 1 } { 5 }$ .   $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } } { x ^ { 2 } }$  
A) $y = \frac { 1 } { 1 - 3 x }$
B)   $y = \frac { x } { 1 - 3 x }$
C)   $y = \frac { x } { 3 - x }$
D)   $y = - \frac { 3 } { 1 - x }$
E)   $y = \frac { x } { 1 + 3 x }$

Quizplus · Accepted Answer

To solve the differential equation, separate variables and integrate: $\int \frac{1}{y^2} \, dy = \int \frac{1}{x^2} \, dx$. This gives $-\frac{1}{y} = -\frac{1}{x} + C$. Solving for $y$, we get $y = \frac{x}{C \cdot x - 1}$. Using the initial condition $y = \frac{1}{2}$ when $x = \frac{1}{5}$, solve for $C$ to find $C = 3$. Thus, the particular solution is $y = \frac{x}{1 - 3x}$.

For the Differential Equation, Find the Particular Solution If y=12y = \frac { 1 } { 2 }y=21​

For the Differential Equation, Find the Particular Solution If $y = \frac { 1 } { 2 }$