Question

4xy dx + (x+1)dy = 0. solve the differential equation​

Answer 1

Answer:

This is a linear first order differential equation. It has a simple procedure to solve for  y(x) .

(x2+1)dydx+4xy=x  

Put it in standard form so that the coefficient of  dydx  is one. Divide both sides by  (x2+1) .

dydx+4x(x2+1)y=x(x2+1)  

Find the integrating factor  ρ(x)  to make the left side match the product rule for derivatives. The right side is solved by direct integration.

ρ(x)=exp(∫4x(x2+1)dx)=exp(2∫2x(x2+1)dx)=e2ln(x2+1)  

ρ(x)=(x2+1)2  

Multiply both sides by the integrating factor  ρ(x) . Reduce the left side to the product rule.

ρ(x)dydx+ρ(x)4x(x2+1)y=ρ(x)x(x2+1)  

ddx(ρ(x)y)=(x2+1)2⋅x(x2+1)  

ddx((x2+1)2⋅y)=(x2+1)2⋅x(x2+1)  

Simplify and integrate both sides with respect to  x .

∫ddx((x2+1)2⋅y)dx=∫(x2+1)xdx  

(x2+1)2⋅y=∫(x2+1)xdx  

(x2+1)2⋅y=14(x2+1)2+C  

Divide by  (x2+1)2  to get  y(x)  alone.

y(x)=14+C(x2+1)−2  

 

y(x)=14+C(x2+1)2  

 

Step-by-step explanation:

Answer 2

Answer:

$8 x \times 2d + 2y$

Hope it's correct thank you please mark me as a brain lies ฅ'ω'ฅ