Question

If the curve satisfying $x\left(x+1\right)y_1 – y = x\left(x+1\right)$ passes through (1, 0), then the value of $ \dfrac{5}{4}y\left( 4 \right)-\log 4$ is ?

Answer 1

Hint: First you need to use the integrator method to find the solution for the differential equation. You need to find the integrator and then multiply it o the both sides of the equation. Then you need to solve the differential equation and finally apply integration to get the equation in y. Now use the point ( 1, 0 ) to get the value of the integration constant. And then finally find the value of y(4) to get the answer to the question.

Complete step by step solution:
Here is the complete step by step solution.
The first step is to divide the equation with x(x + 1). Therefore, we get
$ \Rightarrow \dfrac{dy}{dx}-\dfrac{y}{x\left( x+1 \right)}=1$ .
Now we use the integrator method to solve the differential equation. The integrator for the equation an be found using
$ i={{e}^{\int{-\dfrac{1}{x\left( x+1 \right)}dx}}}$
$ i={{e}^{\int{-\dfrac{1}{x}+\dfrac{1}{x+1}dx}}}={{e}^{\ln \left( \dfrac{1}{x}+1 \right)}}=\dfrac{1}{x}+1$
Now we have to multiply the integrator to both sides of the equation. We get
$ \Rightarrow \left( \dfrac{x+1}{x} \right)\dfrac{dy}{dx}-\left( \dfrac{x+1}{x} \right)\dfrac{y}{x\left( x+1 \right)}=\left( \dfrac{x+1}{x} \right)$
$ \Rightarrow \left( \dfrac{x+1}{x} \right)\dfrac{dy}{dx}-\dfrac{d\left( \dfrac{x+1}{x} \right)}{dx}y=\left( \dfrac{x+1}{x} \right)$
Now we use the formula
$ f\dfrac{dg}{dx}+g\dfrac{df}{dx}=\dfrac{dfg}{dx}$
Therefore, we get
$ \Rightarrow \dfrac{d\left( \dfrac{x+1}{x} \right)y}{dx}=\left( \dfrac{x+1}{x} \right)$
$ \Rightarrow \left( \dfrac{x+1}{x} \right)y=\int{\left( \dfrac{x+1}{x} \right)}dx$
$ \Rightarrow \left( \dfrac{x+1}{x} \right)y=\log x+x+c$
Now we use the point (1,0) to find the value of c
$ \Rightarrow 0=1+c$
Therefore, c = -1.
Now we find y(4)
$ \Rightarrow \dfrac{5}{4}y=\log 4+4-1$
$ \Rightarrow \dfrac{5}{4}y-\log 4=3$

Therefore the answer is 3.

Note: You need to know how to find the focus of a parabola. Also it is important to remember the formulas for the normal and tangent for different shapes, such as parabola, circle, hyperbola, and also the ellipse.