Question
Question: If the curve satisfying \(x\left(x+1\right)y_1 – y = x\left(x+1\right)\) passes through (1, 0), then...
If the curve satisfying x(x+1)y1–y=x(x+1) passes through (1, 0), then the value of 45y(4)−log4 is ?
Solution
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
⇒dxdy−x(x+1)y=1 .
Now we use the integrator method to solve the differential equation. The integrator for the equation an be found using
i=e∫−x(x+1)1dx
i=e∫−x1+x+11dx=eln(x1+1)=x1+1
Now we have to multiply the integrator to both sides of the equation. We get
⇒(xx+1)dxdy−(xx+1)x(x+1)y=(xx+1)
⇒(xx+1)dxdy−dxd(xx+1)y=(xx+1)
Now we use the formula
fdxdg+gdxdf=dxdfg
Therefore, we get
⇒dxd(xx+1)y=(xx+1)
⇒(xx+1)y=∫(xx+1)dx
⇒(xx+1)y=logx+x+c
Now we use the point (1,0) to find the value of c
⇒0=1+c
Therefore, c = -1.
Now we find y(4)
⇒45y=log4+4−1
⇒45y−log4=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.