Question

solve the initial value problem:

$$\frac{dy}{dx}+5xy = 6x,\\ y(0)=2$$

y =

Explanation:

Step1: Identify the integrating factor

For the linear differential equation \(\frac{dy}{dx}+P(x)y = Q(x)\), here \(P(x)=5x\) and \(Q(x) = 6x\).
The integrating factor \(\mu(x)=e^{\int P(x)dx}\).
\(\int5x dx=\frac{5x^{2}}{2}\), so \(\mu(x)=e^{\frac{5x^{2}}{2}}\).

Step2: Multiply the differential equation by the integrating factor

Multiply \(\frac{dy}{dx}+5xy = 6x\) by \(e^{\frac{5x^{2}}{2}}\):
\(e^{\frac{5x^{2}}{2}}\frac{dy}{dx}+5xe^{\frac{5x^{2}}{2}}y = 6xe^{\frac{5x^{2}}{2}}\).
The left - hand side is the derivative of \(y\mu(x)\) by the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (where \(u = y\) and \(v=e^{\frac{5x^{2}}{2}}\)).
So \(\frac{d}{dx}(ye^{\frac{5x^{2}}{2}})=6xe^{\frac{5x^{2}}{2}}\).

Step3: Integrate both sides

Integrate \(\frac{d}{dx}(ye^{\frac{5x^{2}}{2}})=6xe^{\frac{5x^{2}}{2}}\) with respect to \(x\).
Let \(u=\frac{5x^{2}}{2}\), then \(du = 5xdx\) and \(6xe^{\frac{5x^{2}}{2}}dx=\frac{6}{5}e^{u}du\).
\(\int6xe^{\frac{5x^{2}}{2}}dx=\frac{6}{5}e^{\frac{5x^{2}}{2}}+C\).
So \(ye^{\frac{5x^{2}}{2}}=\frac{6}{5}e^{\frac{5x^{2}}{2}}+C\).

Step4: Solve for \(y\)

\(y=\frac{6}{5}+Ce^{-\frac{5x^{2}}{2}}\).

Step5: Use the initial condition \(y(0) = 2\)

Substitute \(x = 0\) and \(y=2\) into \(y=\frac{6}{5}+Ce^{-\frac{5x^{2}}{2}}\).
\(2=\frac{6}{5}+C\), then \(C=2-\frac{6}{5}=\frac{4}{5}\).

Answer:

\(y=\frac{6}{5}+\frac{4}{5}e^{-\frac{5x^{2}}{2}}\)