Hello, I'm solving this equation for y, but the computer system says it's wrong. Any ideas?
Equations:
\(\displaystyle \dfrac{dy}{dx}+P(x)y=Q(x)\) get function into this form before using the equation below this.
\(\displaystyle \displaystyle I(x)=e^{\int P(x)}\)
\(\displaystyle \displaystyle \int udv = uv-\int vdu\)
Work:
1. \(\displaystyle xy'-2y=x^2\)
2. \(\displaystyle \dfrac{dy}{dx}-2y=\dfrac{x^2}{x}\) divide x from the left and change y' into dy/dx
3. \(\displaystyle \displaystyle \dfrac{dy}{dx}-2y=x\) simplify right side
4. \(\displaystyle P(x)=-2\) identify P(x)
5. \(\displaystyle \displaystyle e^{\int -2 dx}\) use second equation and plug in P(x)
6. \(\displaystyle \displaystyle e^{-2x}\) take integral
7. \(\displaystyle e^{-2x}\dfrac{dy}{dx}-2ye^{-2x}=xe^{-2x}\) multiply \(\displaystyle e^{-2x}\) through equation from line 3
8. \(\displaystyle \displaystyle \dfrac{d}{dx}ye^{-2x}= \int xe^{-2x}\) write the left side as the derivative of the product of line 7
integration-by-parts since we have a variable \(\displaystyle x\) which is a product of another function (\(\displaystyle e^{-2x}\))
10. \(\displaystyle \displaystyle ye^{-2x}=-\dfrac{1}{2}xe^{-2x}-\int-\dfrac{1}{2}e^{-2x}dx\) the right hand-side: \(\displaystyle \displaystyle uv-\int vdu\) take the integral of \(\displaystyle -\dfrac{1}{2}e^{-2x}\)
11. \(\displaystyle \displaystyle ye^{-2x}=-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}+C\)
12. \(\displaystyle \dfrac{ye^{-2x}}{e^{-2x}}=\dfrac{-\dfrac{1}{2}xe^{-2x}}{e^{-2x}}-\dfrac{\dfrac{1}{4}e^{-2x}dx}{e^{-2x}}+C\) get y alone
13. \(\displaystyle y=-\dfrac{1}{2}x-\dfrac{1}{4}+Ce^{-2x}\)
Equations:
\(\displaystyle \dfrac{dy}{dx}+P(x)y=Q(x)\) get function into this form before using the equation below this.
\(\displaystyle \displaystyle I(x)=e^{\int P(x)}\)
\(\displaystyle \displaystyle \int udv = uv-\int vdu\)
Work:
1. \(\displaystyle xy'-2y=x^2\)
2. \(\displaystyle \dfrac{dy}{dx}-2y=\dfrac{x^2}{x}\) divide x from the left and change y' into dy/dx
3. \(\displaystyle \displaystyle \dfrac{dy}{dx}-2y=x\) simplify right side
4. \(\displaystyle P(x)=-2\) identify P(x)
5. \(\displaystyle \displaystyle e^{\int -2 dx}\) use second equation and plug in P(x)
6. \(\displaystyle \displaystyle e^{-2x}\) take integral
7. \(\displaystyle e^{-2x}\dfrac{dy}{dx}-2ye^{-2x}=xe^{-2x}\) multiply \(\displaystyle e^{-2x}\) through equation from line 3
8. \(\displaystyle \displaystyle \dfrac{d}{dx}ye^{-2x}= \int xe^{-2x}\) write the left side as the derivative of the product of line 7
integration-by-parts since we have a variable \(\displaystyle x\) which is a product of another function (\(\displaystyle e^{-2x}\))
\(\displaystyle u=x\) | \(\displaystyle dv=e^{-2x}dx\) |
\(\displaystyle du=dx\) | \(\displaystyle v=-\dfrac{1}{2}e^{-2x}\) |
10. \(\displaystyle \displaystyle ye^{-2x}=-\dfrac{1}{2}xe^{-2x}-\int-\dfrac{1}{2}e^{-2x}dx\) the right hand-side: \(\displaystyle \displaystyle uv-\int vdu\) take the integral of \(\displaystyle -\dfrac{1}{2}e^{-2x}\)
11. \(\displaystyle \displaystyle ye^{-2x}=-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}+C\)
12. \(\displaystyle \dfrac{ye^{-2x}}{e^{-2x}}=\dfrac{-\dfrac{1}{2}xe^{-2x}}{e^{-2x}}-\dfrac{\dfrac{1}{4}e^{-2x}dx}{e^{-2x}}+C\) get y alone
13. \(\displaystyle y=-\dfrac{1}{2}x-\dfrac{1}{4}+Ce^{-2x}\)