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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\mathbb{R}$ và ${f}'\left( x \right)-f\left( x \right)=\left( x+1 \right){{e}^{3x}},$ với mọi $x\in \mathbb{R}$. Biết $f\left( 0 \right)=\dfrac{5}{4}$, giá trị $f\left( 1 \right)$ bằng
A. $\dfrac{5}{4}{{e}^{3}}+e$.
B. $\dfrac{3}{4}{{e}^{3}}-e$.
C. $\dfrac{3}{4}{{e}^{3}}+e$.
D. $\dfrac{5}{4}{{e}^{3}}-e$.
A. $\dfrac{5}{4}{{e}^{3}}+e$.
B. $\dfrac{3}{4}{{e}^{3}}-e$.
C. $\dfrac{3}{4}{{e}^{3}}+e$.
D. $\dfrac{5}{4}{{e}^{3}}-e$.
Ta có: ${f}'\left( x \right)-f\left( x \right)=\left( x+1 \right){{e}^{3x}}\Leftrightarrow {{e}^{-x}}{f}'\left( x \right)-{{e}^{-x}}f\left( x \right)=\left( x+1 \right){{e}^{2x}}\Leftrightarrow {{\left( {{e}^{-x}}f\left( x \right) \right)}^{\prime }}=\left( x+1 \right){{e}^{2x}}$
Khi đó: ${{e}^{-x}}f\left( x \right)=\int{\left( x+1 \right){{e}^{2x}}}dx=\dfrac{1}{2}\left( x+1 \right){{e}^{2x}}-\dfrac{1}{4}{{e}^{2x}}+C$
Do $f\left( 0 \right)=\dfrac{5}{4}$ nên: $\dfrac{1}{4}+C=\dfrac{5}{4}\Leftrightarrow C=1\Rightarrow f\left( x \right)=\dfrac{1}{2}\left( x+1 \right){{e}^{3x}}-\dfrac{1}{4}{{e}^{3x}}+{{e}^{x}}$
Vậy $f\left( 1 \right)=\dfrac{3}{4}{{e}^{3}}+e$.
Khi đó: ${{e}^{-x}}f\left( x \right)=\int{\left( x+1 \right){{e}^{2x}}}dx=\dfrac{1}{2}\left( x+1 \right){{e}^{2x}}-\dfrac{1}{4}{{e}^{2x}}+C$
Do $f\left( 0 \right)=\dfrac{5}{4}$ nên: $\dfrac{1}{4}+C=\dfrac{5}{4}\Leftrightarrow C=1\Rightarrow f\left( x \right)=\dfrac{1}{2}\left( x+1 \right){{e}^{3x}}-\dfrac{1}{4}{{e}^{3x}}+{{e}^{x}}$
Vậy $f\left( 1 \right)=\dfrac{3}{4}{{e}^{3}}+e$.
Đáp án C.