The equation3y′=3xy+2x2+2y2x2, (∗)can be written in the form y′=f(y/x), i.e., it is homogeneous, so we can use the substitution
u=y/x to obtain a separable equation with dependent variable u=u(x). Introducing this substitution and using the fact that y′=xu′+u we can write (∗) asy′=xu′+u=f(u)where f(u)=Separating variables we can write the equation in the formg(u)du=dxxwhere g(u)= . An implicit general solution with dependent variable u can be written in the formln(x)−=CTransforming u=y/x back into the variables x and y and using the initial condition y(1)=1 we findC=Finally solve for y to obtain the explicit solution of the initial value problemy=
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Question : Draw a triangle ABC. Next, create a line through point C that is perpendicular to . Label the intersection between the perpendicular line and as point E. Take a screenshot of the triangle with CE, save it, and insert the image in the space below.
We know that having three sides, three angles, and three vertices, a triangle is a closed, two-dimensional shape. A polygon also includes a triangle.
A line is a straight, one-dimensional figure in geometry that is infinitely long in both directions and has no thickness.
The definition of a perpendicular line in geometry is a pair of lines that meet or intersect at right angles (90°). The Latin word "perpendicularis," which denotes a plumb line, is where the word "perpendicular" first appeared. Two lines, AB and CD, can be written as AB⊥CD if they are perpendicular to one another. The perpendicular nature of the lines is denoted by the symbol.
Here, we have to draw a triangle. Then we have to draw a line CE which intersects AB at point E.
Image is attached below.
Learn more about perpendicular lines here -
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