Solve the homogeneous equations in Exercises 5–10. First put t...

Question

Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.

(x.exp(y/x) + y)dx - x dy = 0

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Solve the differential equation by reducing it to a homogeneous equation. You may leave your answer in implicit form. Parentheses of x multiplied by e to the power of 2y divided by x plus YDX minus XDY is equal to 0. OK. So, it appears for this particular prompt, we're asked to ultimately solve the differential equation that is provided to us by the prom itself, and we're asked to solve for this specific differential equation by reducing it to a homoogeneous equation. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to rewrite our provided differential equation. So we need to take parentheses of x multiplied by e to the power of 2y divided by x. Plusy multiplied by d x minus x multiplied by d y is equal to 0, and when we rewrite this expression, we can write parentheses of x multiplied by e to the power of 2 y divided by x plus y multiplied by dx will be equal to x multiplied by dy. Therefore, we could write that Dy by dx will be equal to x multiplied by e to the power of 2y divided by x plus y divided by x will be equal to e to the power of 2y divided by x plus y divided by x. OK. Our second step now is we need to make our substitution, so we need to recall and use the substitution method. So we will state that V will be equal to Y divided by X. Then that will mean as a result, Y is equal to VX. Which will mean we could write that DY by DX is equal to V plus XDV by DX. And we need to substitute into our differential equation, so we could write V plus X multiplied by DV by DX will be equal to E to the power of 2V plus V, and we now need to subtract V from both sides, and that will leave us with DV by DX is equal to E to the power of 2V. Our third step now is we need to separate our variables, so we need to take DV divided by E to the power of 2V, which will be equal to DX divided by X. And we need to rewrite the left side, and when we do that, you will get E to the power of -2V. So once again we have E to the power of -2 VDV. And this will be equal to DX divided by X. Our fourth step now is we need to integrate both sides, so we take the integral of e to the power of -2 V DV is equal to the integral of the x divided by x. Fantastic. And we now need to calculate each integral, so we need to note that the integral of e to the power of -2VDV will be equal to 1/2 multiplied by e to the power of 2V plus C subscript 1, where C denotes some constant variable. And the integral of dx divided by X will be equal to the natural log of the absolute value of X plus C subscript 2. So therefore, we can go ahead and write that 1/2 multiplied by e to the power of -2V plus C1 is equal to the natural log of the absolute value of X plus C2. Which means that 1/2 multiplied by e to the power of -2V. Will be equal to the natural log of the absolute value of X plus C subscript 3, which I'll call C3 for short. And let's make a note off to the side that C3 is equal to C2 minus C1. So at this stage, so this will be for step 5, we need to substitute back in our value of V is equal toy divided by X. So when we substitute this value of v back into our expression, we will write 1/2 multiplied by e to the power of 2y divided by x will be equal to the natural log of the absolute value of x plus c3. We now need to multiply both sides by -2. And when we do that, we will find that e to the power of -2Y divided by X will be equal to -2 multiplied by the natural log of the absolute value of x plus C. And we need to note that C is equal to -2 multiplied by C3. And when we rearrange our expression, we will be able to write that e to the power of 2 y divided by x plus 2 multiplied by the natural log of the absolute value of x will be equal to c, and this is our correct and final answer. Hooray, we did it. So going back to look at the problem itself, we could safely say without a doubt that our final answer for the differential equation is once again. E to the power of -2y divided by x plus 2 multiplied by the natural log of the absolute value of x, which is equal to c, and that's it. That is our correct and final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.(x.exp(y/x) + y)dx - x dy = 0

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Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.
(x.exp(y/x) + y)dx - x dy = 0