Let y = y(x) be the solution of the differential equation (x² + 1)y′ − 2xy = (x⁴ + 2x² + 1)cosx, y(0) = 1. Then ³∫₋₃ y(x) dx is:

 Question

Let $y = y(x)$be the solution of the differential equation

$$(x^2+1)y^{\mathrm{\prime }}-2xy=(x^4+2x^2+1)\cos x,$$

with $y(0) = 1$.
Then evaluate:

$${\int }_{-3}^{3}y(x)dx$$

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Question Image

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Short Solution

1. Rewrite the ODE in standard linear form:

   $$y^{\mathrm{\prime }}-\frac{2x}{x^2+1}y=\frac{(x^4+2x^2+1)\cos x}{x^2+\mathrm{1}}$$

2. Find the integrating factor:

   $$\mu (x)=e^{\int -\frac{2x}{x^2+1}dx}=e^{-\ln (x^2+1)}=\frac{1}{x^2+\mathrm{1}}$$

3. Multiply the equation by $\mu(x)$:

   $$\frac{d}{dx}\left(\frac{y}{x^2+1}\right)=\cos x$$

4. Integrate:

   $$\frac{y}{x^2+1}=\sin x+C$$

5. Apply $y(0)=1$:

   $$\frac{1}{1}=0+C⟹C=1$$

6. Hence:

   $$y(x)=(x^2+1)(\sin x+1)$$

7. Compute the definite integral:

   $$I={\int }_{-3}^{3}y(x)dx={\int }_{-3}^3(x^2+1)(\sin x+1)dx$$

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Image Solution

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Conclusion

Break the integral:

$$I={\int }_{-3}^3(x^2+1)\sin xdx+{\int }_{-3}^3(x^2+1)dx$$

• First term: $(x^{2}+1)\sin x$ is an odd function (because $x^{2}+1$ is even and $\sin x$ is odd)), so its integral over $[-3,3]$ is 0.

• Second term:

$${\int }_{-3}^3(x^2+1)dx={[\frac{x^3}{3}+x]}_{-3}^3=(9+3)-(-9-3)=12-(-12)=24$$

✅ Final Answer:

$$\boxed{{\displaystyle {\int }_{-3}^{3}y(x)dx=\mathrm{24}}}$$

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Video Solution